6 TRIANGLES

Exercise 6.1
Q.1. Fill in the blanks using the correct word given in brackets:
(i) All circles are ______. (congruent, similar)
(ii) All squares are ______. (similar, congruent)
(iii) All ______ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are ______ and (b) their corresponding
sides are ______. (equal, proportional)
Solution. (i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.
Q.2. Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
Solution. (i) Two different examples of similar figures are:
(a) Two twenty-rupees notes.
(b) Two two-rupees coins.
(ii) Two different examples of non-similar figures are:
(a) One-rupee coin and a five rupees coin.
(b) One-rupee note and ten rupees note.
Q.3. State whether the following quadrilaterals are similar or not:

1
Solution. The two quadrilaterals in the given figure are not similar because their
corresponding angles are not equal.

Exercise 6.2
Q.1. In figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution. (i) in ΔABC, DE || BC (Given)

AD AE
Therefore, = [By using Basic Proportionality Theorem]
DB EC
1.5 1
⇒ =
3 EC
3
⇒ Σ EC =
1.5
3 × 10
EC = = 2 cm
15
Hence, EC = 2 cm
(ii) In ΔABC,
DE || BC (Given)
AD AE
Therefore, = [By using Basic Proportionality Theorem]
DB EC
AD 1.8
⇒ =
7.2 5.4
1.8 × 7.2 18 72 10 24
⇒ AD = = × × =
5.4 10 10 54 10
⇒ AD = 2.4
Hence, AD = 2.4 cm.

2
Q.2. E and F are points on the sides PQ and PR respectively of a ΔPQR . For each of
the following cases, state whether EF || OR :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution. In ΔPQR , E and F are two points on side PQ and PR respectively

(i) PE = 3.9 cm, EQ = 3 cm (Given)

PF = 3.6 cm, FR = 2.4 cm (Given)
PE 3.9 39 13
Therefore, = = = = 1.3 (By using Basic Proportionality Theorem)
EQ 3 30 10
PF 3.6 36 3
And, = = = = 1.5
FR 2.4 24 2
PE PF
So, ≠
EQ FR
Hence, EF is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
PE 4 40 8
Therefore, = = = (By using Basic Proportionality Theorem)
QE 4.5 45 9
PF 8
And, =
RF 9
PE PF
So, =
QE RF
Hence, EF is parallel to QR
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm (Given)
Here, EQ = PQ − PE = 1.28 − 0.18 = 1.10 cm
And, FR = PR − PF = 2.56 − 0.36 = 2.20 cm

3
 PE 0.18 18 9
So, = = = ...(i)
EQ 1.10 110 55
PE 0.36 36 9
And, = = = ...(ii)
FR 2.20 220 55
PE PF
Therefore, =
EQ FR
Hence, EF is parallel to QR.
Q.3. In figure, LM || CB and LN || CD, prove that
AM AN
= .
AB AD
Solution. In ΔABC, we have
ML || BC (Given)

AM AL
⇒ = (By using Basic Proportionality theorem) ...(i)
MB LC
In ΔACD, we have
LN || DC (Given)
AN AL
⇒ = (By using Basic Proportionality Theorem) ...(ii)
ND LC
From equation (i) and equation (ii), we get
AM AN
=
MB ND
MB ND
or, =
AM AN
MB ND
or, +1 = +1 (Adding 1 in both sides)
AM AN
MB + AM ND + AN
or, =
AM AN
AB AD
or, =
AM AN
AM AN
So, =
AB AD

4
 BF BE
Q.4. In figure DE || AC, and DF || AE, prove that = .
FE EC

Solution. In ΔABC, DE || AC (Given)

BD BE
Therefore, = ...(i) (By using Basic Proportionality Theorem)
DA EC
In ΔABE, DF || AE (Given)
BD BF
Therefore, = ...(ii) (By using Basic Proportionality Theorem)
DA FE
From equation (i) and equation (ii), we have
BE BF
=
EC FE
Q.5. In figure DE || OQ and DF || OR. Show that EF || QR.
Solution. Given: DE || OQ and DF || OR
To prove: s EF || QR.
Proof: In ΔPQO, DE || OQ (Given)

PD PE
Therefore, = ...(i) (By using Basic Proportionality Theorem)
DO EQ
In ΔPOR, DF || OR
PD PF
Therefore, = ...(ii) (By using Basic Proportionality Theorem)
DO FR
From equation (i) and equation (ii), we have
PE PF
=
EQ FR
In ΔPQR, by using converse of Basic Proportionality Theorem.
EF is parallel to QR.

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Q.6. In figure, A, B and C are points on OP, OQ and OR
respectively such that AB || PQ and AC || PR.
Show that BC || QR.
Solution. Given: ΔPQR, A, B and C are points on OP, OQ and
OR respectively such that
AB || PQ and AC || PR
To prove: BC || QR.
Proof: In ΔOPQ, we have

AB || PQ (Given)
OA OB
Therefore, = ...(i) (By using Basic Proportionality Theorem)
AP BQ
In ΔOPR, we have
AC || PR (Given)
OA OC
Therefore, = ...(ii) (By using Basic Proportionality Theorem)
AP CR
From equation (i) and (ii), we have
OB OC
=
BQ CR
Hence, in ΔOQR, BC||QR, (By Converse of Basic Proportionality Theorem)
Q.7. Using Basic Proportionality Theorem, prove that a line drawn through the mid-
point of one side of a triangle parallel to another side bisects the third side.
(Recall that you have proved it in class IX).
Solution. Given: ΔABC, D is the mid point of AB, i.e., AD = DB.
A line parallel to BC intersects AC at E as shown in figure, i.e.,
DE || BC.
To prove: E is the mid point of AC.
Proof: D is the mid point of AB.
Therefore, AD = DB
AD
⇒ =1 ...(i)
BD
In ΔABC, DE || BC
AD AE
Therefore, = (By Basic Proportionality Theorem)
DB EC

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 AE
⇒ 1= [From equation (i)]
EC
Therefore, AE = EC
Hence, E is the mid-point of AC, proved.
Q.8. Using converse of Basic Proportionality Theorem, prove that the line joining
the mid-points of any two sides of a triangle is parallel to the third side. (Recall
that you have done it in class IX).
Solution. Given: ΔABC, D and E are the mid points of AB and AC respectively such
that AD = BD and AE = EC,
To prove: DE || BC
Proof: D is the mid-point of AB(Given)
Therefore, AD = BD
AD
⇒ =1 ...(i)
BD
Aslo, E is the mid-point of AC (Given)
Therefore, AE = EC
AE
=1 ...(ii)
EC
AD AE
From equation (i) and equation (ii), we have =
BD EC
Hence, DE || BC , (By converse of Basic Proportionality Theorem)
Q.9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other
AO CO
at the point O. Show that = .
BO DO
Solution. Given: ABCD is a trapezium in which AB || DC, diagonals AC and BD
intersect each other at O.
AO CO
To prove: =
BO DO
Construction: Through O, draw EO || DC || AB
Proof: In ΔADC, we have
EO || DC (By construction)
AO AE
Therefore, = ...(i) (By using Basic Proportionality Theorem)
CO DE
In ΔDBA , we have
EO || AB (By construction)

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 DE DO
= ...(ii) (By using Basic Proportionality Theorem)
EA BO
From equation (i) and equation (ii), we get
AO BO
=
CO DO
AO CO
or, =
OB DO
Q.10. The diagonals of a quadrilateral ABCD intersect each other at the point O
AO CO
such that = . Show that ABCD is a trapezium.
BO DO
Solution. Given: Quadrilateral ABCD, diagonals AC and BD intersects each other at O
AO CO
such that = .
BO DO
To Prove: ABCD is trapezium.
Construction: Through O, draw line EO, where EO || AB which meets AD at E.
Proof: In ΔDAB, we have
EO || AB
DE DO
Therefore, = ...(i) (By using Basic Proportionality Theorem)
EA OB
AO CO
Also, = (Given)
BO DO
AO BO
or, =
CO DO
CO DO
or, =
AO BO
DO CO
⇒ = ...(ii)
OB AO
From equation (i) and equation (ii), we get
DE CO
=
EA AO
Therefore, By using converse of Basic Proportionality Theorem, EO || DC also
EO||AB ⇒ AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD.

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 Exercise 6.3
Q.1. State which pairs of triangles in Figure are similar. Write the similarity
criterion used by you for answering the question and also write the pairs of
similar triangles in the symbolic form:

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Solution. (i) In ΔABC and ΔPQR, We have
∠A = ∠P = 60° (Given)
∠B = ∠Q = 80° (Given)
∠C = ∠R = 40° (Given)
Therefore, ΔABC ~ ΔPQR (AAA similarity criterion)
In ΔABC and ΔPQR, We have
(ii) AB = 2, BC = 2.5, AC = 3, QR = 4, PR = 5, PQ = 6
AB 2 1
= = ...(i)
RQ 4 2
AC 3 1
= = ...(ii)
PQ 6 2
BC 2.5 1
= = ...(iii)
PR 5 2
From equation (i), equation (ii) and equation (iii), we have
AB AC BC 1
= = =
RQ PQ PR 2
Therefore, ΔABC ~ ΔQRP (SSS similarity criterion)
(iii) In ΔLMP and ΔDEF , We have
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP 2 1
= =
DE 4 2

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 PL 3 1
= =
DF 6 2
LM 2.7 27
= =
EF 5 50
MP PL LM
Here, = ≠ .
DE DF EF
Hence, ΔLMP and ΔDEF are not similar.
(iv) In ΔMNL and ΔPQR, We have
MN = 3, ML = 5, QR = 10, PQ = 5, ∠M = 70°, ∠A = 70°
ML 5 1
Here, = =
QR 10 2
Aslo, ∠M = ∠Q = 70°
MN 3 1
And, = =
PQ 6 2
Hence, ΔMNL ~ ΔQPR (By SAS similarity criterion)
(v) In ΔABC and ΔDEF, We have
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
AB 2.5 1
Here, = =
DF 5 2
BC 3 1
And, = =
EF 6 2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.
(vi) In ΔDEF , We have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° − 70° − 80°
⇒ ∠F = 30°
In ΔPQR, We have
∠P + ∠Q + ∠R = 180 (Sum of Angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° − 80° − 30°
⇒ ∠P = 70°
In ΔDEF and ΔPQR, We have
∠D = ∠P = 70°

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 ∠E = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR (By AAA similarity criterion)
Q.2. In Figure, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find
∠DOC, ∠DCO and ∠OAB.
Solution. Given that: ∠BOC = 125°
∠CDO = 70°
Therefore, ∠DOC + ∠COB = 180° (Linear pair axiom)
⇒ ∠DOC + 125° = 180°
⇒ ∠DOC = 180° − 125°
⇒ ∠DOC = 55°
⇒ ∠DOC = ∠AOB = 55° (Vertically opposite angles)
But ΔODC ~ ΔOBA (Given)
Therefore, ∠D = ∠B = 70°
In ΔDOC, ∠D + ∠DOC + ∠C = 180°
⇒ 70° + 55° + ∠C = 180°
⇒ ∠C = 180° − 70° − 55°
⇒ ∠C = 55°
∠C = ∠A = 55° [Since, ΔODC ~ ΔOBA]
Hence, ∠DOC = 55 °
∠DCA = 55 °
∠OAB = 55 °
Q.3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other
at the point O. Using a similarity criterion for two triangles, show that
OA OB
= .
OC OD
Solution. Given: Trapezium ABCD, AB || CD and diagonals AC and BD intersects each
other at O.
AO OB
To prove: =
OC OD
Proof: In ΔOAB and ΔOCD, We have
AB || CD
Here, ∠OAB = ∠OCD (alternate angles)
∠AOB = ∠COD (vertically opposite angle)
And, ∠OBA = ∠ODC (alternate angles)

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 Therefore, ΔOAB ~ ΔOCD (AAA Similarity, Criteion)
DO OC
⇒ = [If two triangles are similar then corresponding sides are proportional)
BO OA
OA BO
Hence, =
OC DO
QR QT
Q.4. In figure = and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
QS PR
Solution. Given that,
QR QT
= and ∠1 = ∠2
QS PR
To prove: ΔPQS ~ ΔTQR
Proof: In ΔPQR, We have
∠1 = ∠2 (Given)
Therefore, PR = PQ (sides opposite to equal angles of a Δ are equal) ...(i)
QR QT
and, = (Given)
QS PR
QR QT
or, = (PR = PQ) ...(ii)
QS QP
QS PQ
⇒ =
QR QT
In ΔPQS and ΔTQR, We havess
QS PQ
=
QR QT
Also, ∠SQP = ∠RQT
⇒ ∠1 = ∠1
Hence, ΔPQS ~ ΔTQR (SSS similarity criterion)
Q.5. S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show
that ΔRPQ ~ ΔRTS.
Solution. In ΔPQR, S and T are points on side PR and QR of ΔPQR such that
∠P = ∠RTS
To prove:S ΔRPQ ~ ΔRTS
Proof: In ΔRPQ and ΔRTS
∠RPQ = ∠RTS (Given)

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 ∠R = ∠R s (Common angle)
Therefore, ΔRPQ ~ ΔRTS (By AA similarity criterion)
Q.6. In figure if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Solution. Given: ΔABC in which ΔABE ≅ ΔACD
To prove: ΔADE ~ ΔABC
Proof: ΔABE ≅ ΔACD (Given)
⇒ AB = AC (by cpct)
And, AE = AD (by cpct)
AB
⇒ =1 ...(i)
AC
AE
=1 ...(ii)
AD
From equation (i) and (ii), We have
AB AD
=
AC AE
In ΔADE and ΔABC, We have
AD AB
=
AE AC
And, ∠A = ∠A (common)
Hence, ΔADE ~ ΔABC (By SAS similarity criterion)
Q.7. In the given figure, altitudes AD and CE of ΔABC intersect each other at the
point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC
Solution. Given: ΔABC, AD ⊥ BC, CE ⊥ AB,
To prove: (i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC
Proof: (i) In ΔAEP and ΔCDP, We have
∠AEP = ∠CDP = 90° ...(i)
And, ∠EPA = ∠DPC ...(ii) (Vertically opposite angle)

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 From equation (i) and equation (ii), we have
∴ ΔAEP ~ ΔCDP (By AA Similarity Criterion)
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB = 90 ° ...(i)
And, ∠ABD = ∠CBE ...(ii)
From equation (i) and equation (ii), We have
ΔABD ~ ΔCBE (By AA similarity criterion)
(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB = 90° ...(i)
∠EAP = ∠DAB ...(ii)
From equation (i) and equation (ii), we have
ΔAEP ~ ΔADB (By AA similarity criterion)
(vi) In ΔPDC and ΔBEC, We have
∠DCP = ∠ECB ...(i) (common)
∠PDC = ∠BEC = 90° ...(ii)
Hence, ΔPDC ~ ΔBEC (By AA similarity criterion)
Q.8. E is a point on the side AD produced of a parallelo-gram
ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Solution. Given: Parallelogram ABCD. Side AD is produced to E and
BE intersects DC at F.
To Prove: ΔABE ~ ΔCFB
Proof: In ΔABE and ΔCFB. , We have
∠BAF = ∠FCB ...(i) (opposite angles of || gm)
∠AEB = ∠CBF ...(ii) (alternate angles)
From equation (i) and equation (ii), we have
ΔABE ~ ΔCFB (By AA similarity criterion)
Q.9. In the given figure, ABC and AMP are two right triangles, right angled at B
and M respectively. Prove that:
(i) ΔABC ~ ΔAMP
CA BC
(ii) = .
PA MP
Solution. Given: ΔABC and ΔAMP are two right angled triangles at B and M
respectively.
To Prove: (i) ΔABC ~ ΔAMP
CA BC
(ii) =
PA MP

15
 Proof: (i) In ΔABC and ΔAMP, We havess
∠A = ∠A (common angle)
∠ABC = ∠AMP = 90° (each 90°)
Therefore, ΔABC ~ ΔAMP (By AA similarity criterion)
(ii) As, ∠ABC = ∠ΣAMP
If two triangle are similar, then corresponding sides are equal
CA BC
Hence, =
PA MP
Q.10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D
and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If
ΔABC ~ ΔFEG, show that:
CD AC
(i) = ss
GH FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Solution. Given: In ΔABC and ΔEFG, CD and GH are bisectors of ∠ACB and ∠EGF
respectively
∠1 = ∠2
And, ∠3 = ∠4
ΔABC ~ ΔFEG
CD AC
To prove: (i) =
GH FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Proof:
(i) Given that, ΔABC ~ ΔFEG
Therefore, ∠A = ∠F; ∠B = ∠E
and ∠C = ∠G
[∵ The corresponding angles are equal]
Consider, ∠C = ∠G [Proved above]
1 1
⇒ ∠C = ∠G
2 2
⇒ ∠2 = ∠4 or ∠1 = ∠3
Now, in ΔACD and ΔFGH , We have
∠A = ∠F [Proved above]

16
 ⇒ ∠2 = ∠4 [Proved above]
Therefore, ΔACD ~ ΔFGH [By AA similarity criterion]
CD AG
Hence, = [Since, corresponding sides are proportional]
GH FG
(ii) In ΔDCB and ΔHGE, We have
∠B = ∠E [Proved above]
And, ∠1 = ∠3 [Proved above]
Hence, ΔDCB ~ ΔHGE [By AA similarity criterion]
(iii) In ΔDCA and ΔHGF, We have
∠A = ∠F and ∠2 = ∠4
Hence, ΔDCA~ΔHGF [By AA similarity criterion]
Q.11. In the given Figure, E is a point on side CB produced of an isosceles triangle
ABC with AB = AC, If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Solution. Given: An isosceles triangle ABC with AB = AC, AC ⊥ BC,
side BC is produced to E. EF ⊥ AC
To prove: ΔABD ~ ΔECF
Proof: [ABC is an isosceles triangle 6]
AB = AC
Therefore, ∠ABC = ∠ACB [Equal sides have equal angles opposite to it]
Or, ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ABD = ∠ECF (Proved above)
∠ADB = ∠EFC = 90°
Hence, ΔABD ~ ΔECF [By AA similarity criterion]
Q.12. Sides AB and BC and median AD of a triangle ABC are respectively
proportional to sides PQ and QR and median PM of ΔPQR (see figure). Show
that ΔABC ~ ΔPQR.
Solution. Given: ΔABC and ΔPQR, AB, BC, and median AD of ΔABC are proportional
to sides PQ, QR and median PM of ΔPQR,
AB BC AD
i.e., = =
PQ QR PM
To Prove: ΔABC ~ ΔPQR

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 AB BC AD
Proof: = =
PQ QR PM

1
BC
AB 2 AD
⇒ = = ...(i)
PQ 1 PM
QR
2
AB BD AD
⇒ = = (D is the mid-point of BC. M is the mid-point of QR)
PQ QM PM
⇒ ΔABD ~ ΔPQM [SSS similarity criterion]
Therefore, ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]

⇒ ∠ABC = ∠PQR,
In ΔABC and ΔPQR
AB BC
= ...(i)
PQ QR
∠ABC = ∠PQR ...(ii)
From equation (i) and equation (ii), we have
ΔABC ~ ΔPQR. [By SAS similarity criterion]
Q.13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show
that CA 2 = CB ⋅ CD.
Solution. Given: ΔABC, D is a point on side BC such that ∠ADC = ∠BAC.
To Prove: CA 2 = BC × CD
Proof: In ΔABC and ΔADC.
∠C = ∠C (Common)
∠BAC = ∠ADC (Given)
Therefore, ΔABC ~ ΔDAC [By AA similarity criterion]
AC BC
Therefore, = [If two triangles are similar corresponding sides are proportional]
DC AC
Hence, CA 2 = CB ⋅ CD
Q.14. Sides AB and AC and median AD of a triangle ABC are respectively
proportional to sides PQ and PR and median PM of another triangle PQR.
Show that ΔABC ~ ΔPQR.

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Solution. Given: Two triangles ABC and PQR in which AD and PM are medians such
AB AC AD
that = =
PQ PR PM

To prove : ΔABC ~ ΔDEF

Construction : Produce AD to E so that AD = DE. Join CE, Similarly produce
PM to N such that PM = MN, also join RN.
Proof : In ΔABD and ΔCDE, We have
AD = DE [By construdtion]
BD = DC [∴ AP is the median]
and, ∠ADB = ∠CDE [Vertically opp. angles]
Therefore, ΔABD ≅ ΔCDE [By SAS Criterion of Congruence]
⇒ AB = CE [CPCT] …(i)
Aslo, in ΔPQM and ΔMNR, We have
PM = MN [By Construction]
QM = MR [∴ PM is the median]
and, ∠PMQ = ∠NMR [Vertically opp. ∠S]
Therefore, ΔPQM = ΔMNR [By SAS Criterion of congruence]
⇒ PQ = RN [CPCT] …(ii)
AB AC AD
Now, = =
PQ PR PM
CE AC AD
⇒ = = ….[From (i) and (ii)]
RN PR PM
CE AC 2AD
⇒ = =
RN PR 2PM
CE AC AE
⇒ = = [∴ 2AD = AE and 2PM = PN]
RN PR PN
∴ ΔACE ~ ΔPRN [By SSS Similarly Condition]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
Therefore, ∠ 1 + ∠ 2 = ∠3 + ∠4

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 ⇒ ∠A = ∠P …….. (iii)
Now, In ΔABC and ΔPQR, We have
AB AC
= [Given]
PQ PR
∠A = ∠P [From (iii)]
Therefore, ΔABC ~ ΔPQR [By SAS Similarly Condition]
Q.15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the
same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution. Length of the vertical pole = 6 m (Given)
Shadow of the pole = 4 m (Given)
Let height of tower = h m
Length of shadow of tower = 28 m (Given)
In ΔABC and ΔDFE
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90
Therefore, ΔABC~ΔDEF (By AA similarity criterion)
AB BC
Therefore, = [If two triangles are similar corresponding sides are proportional]
DF EF
6 4
Therefore, =
h 28
6 × 28
⇒ h=
4
⇒ h = 6×7
Or, h = 42 m
Hence, height of the tower = 42 m.
Q.16. If AD and PM are medians of triangles ABC and PQR, respectively where
AB AD
ΔABC ~ ΔPQR, prove that = .
PQ PM
Solution. Given: AD and PM are median the medians of ΔABC and ΔPQR respectively
and ΔABC~ΔPQR
AB AD
To Prove: =
PQ PM
Proof: ΔABC ~ ΔPQR (Given)
AB BC AC
Therefore, = = [If two triangles are similar corresponding sides are proportional]
PQ QR PR
∠A = ∠P

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 ∠B = ∠Q
and, ∠C = ∠R
As D is the mid-point of BC
1
So, BD = DC = BC ...(i)
2
Also, M is the mid-point of QR.
1
So, QM = MR = QR ...(ii)
2
AB BC
⇒ = [∴ ΔABC ~ ΔPQR]
PQ QR
AB 2BD
⇒ = [From equation (i) and equation (ii)]
PQ 2QM
AB BD
⇒ =
PQ QM
And, ∠ABD = ∠PQM (Given)
Therefore, ΔABC ~ ΔPQM (By SAS similarity criterion)
AB AD
Hence, = [If two triangles are similar corresponding sides are proportional]
PQ PM

Exercise 6.4
Q.1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm 2 and 121 cm 2 . If
EF = 15.4 cm, find BC.
Solution. It is given that,
Area of ΔABC = 64 cm 2
Area of ΔDEF = 121 cm 2
EF = 15.4 cm
And, ΔABC ~ ΔDEF
Area of ΔABC AB2
Therefore, =
Area of ΔDEF DE 2
AC2 BC2
= = ...(i)
DF2 EF2
[If two triangles are similar, ratio of their areas are equal to square of the ratio of
their corresponding sides]

21
 64 BC 2
Therefore, =
121 EF2
2 2
⎛ 8 ⎞ ⎛ BC ⎞
⇒ ⎜ ⎟ =⎜ ⎟
⎝ 11 ⎠ ⎝ 15.4 ⎠
8 BC
⇒⇒ =
11 15.4
8 × 15.4
⇒ BC =
11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm
Q.2. Diagonals of a trapezium ABCD with AB || DC interesect each other at the
point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Solution. ABCD is a trapezium with AB || DC. Diagonals
AC and BD intersect each other at the point O.
In ΔAOB and ΔCOD, We have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
Therefore, ΔAOB ~ ΔCOD [By AAA similarity criterion]
Area of (ΔAOB) AB 2 [If two triangles are similar then ratio of
Now, =
Area of (ΔCOD) CD 2 their areas are equal to the square of the
ratio of their corresponding sides]
(2CD) 2
= [∴ AB = 2CD]
CD 2
Area of (ΔAOB) 4CD 2 4
Therefore, = =
Area of (ΔCOD) CD 2 1
Hence, required ratio of area of ΔAOB and ΔCOD = 4 :1
Q.3. In the figure, ABC and DBC are two triangles on the
same base BC. If AD intersects BC at O, show that
ar (ABC) AO
= .
ar (DBC) DO

Solution. Given: ABC and DBC are the triangles on the same base BC. AD intersects
BC at O.

22
 Area of (ΔABC) AO
To Prove: =
Area of (ΔDBC) DO
Construction: Draw AL ⊥ BC, DM ⊥ BC
Proof: In ΔALO and ΔDMO, We have
∠1 = ∠2 (Vertically opposite angles)
∠ALO ∠L = ∠DMO = 90°
Therefore, ΔALO ~ ΔDMO (By AA similarity criterion)
AL AO
⇒ = [If two triangles are similar ...(i)
DM DO
corresponding sides are proportional]

1
× BC × AL
Area of (ΔABC) 2 1
Now, = [Area of triangle = × Base × Altitude]
Area of (ΔDBC) 1 × BC × DM 2
2
From equation (i), We have
Area of (ΔABC) AL
=
Area of (ΔDBC) DM
Area of (ΔABC) AL AO
Hence, = =
Area of (ΔDBC) DM DO
Q.4. If the areas of two similar triangles are equal, prove that they are congruent.
Solution. Given: ΔABC and ΔPQR are similar and equal in area.
To Prove: ΔABC ≅ ΔPQR
Proof: Since, ΔABC ~ ΔPQR
Area of ( ΔABC ) BC2
Therefore, =
Area of ( ΔPQR ) QR 2

BC 2
⇒ =1 [Since, Area (ΔABC) = (ΔPQR)]
QR 2
⇒ BC2 = QR2
⇒ BC = QR
Similarly, we can prove that,
AB = PQ and AC = PR
Therefore, ΔABC ≅ ΔPQR (By SSS Criterion of congruence)

23
Q.5. D, E and F are respectively the mid-points of the sides AB, BC and CA of
ΔABC. Determine the ratio of the areas of ΔDEF and ΔABC.
Solution. Given: D, E and F are the mid-points of the sides AB, BC and CA respectively
of the ΔABC.
To find: area (ΔDEF) and area (ΔABC)
Proof: In ΔABC, We have
F is the mid-point of AB (Given)
E is the mid-point of AC (Given)
So, by the mid-point theorem, We have
1
FE || BC and FE = BC
2
1
⇒ FE || BD and FE = BD [∵ BD = BC]
2
∴ BDEF is a || gm. [∵ Opp. sides of a || gm are parallel and || equal]
Similarly in ΔFBD and ΔDEF, We have
FB = DE (Opposite sides of || gm BDEF)
FD = FD (Common)
BD = FE (Opposite sides of || gm BDEF)
∴ ΔFBD ≅ ΔDEF (SSS Congruence Theorem)
Similarly, we can prove that:
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
If triangles are congruent, then they are equal in area.
So, area (ΔFBD) = area (ΔDEF) ...(i)
area (ΔAFE) = area (ΔDEF) ...(ii)
and, area (ΔEDC) = area (ΔDEF) ...(iii)
Now, area (ΔABC) = area (ΔFBD) + area (ΔDEF)
+ area (ΔAFE) + area (ΔEDC) ...(iv)
area (ΔABC) = area (ΔDEF) + area (ΔDEF) + area (ΔDEF) + area (ΔDEF)
1
⇒ area (ΔDEF) = area (ΔABC) [From (i), (ii) and (iii)]
4
area (ΔDEF) 1
⇒ =
area (ΔABC) 4
Hence, area (ΔDEF) : area (ΔABC) = 1 : 4

24
Q.6. Prove that the ratio of the areas of two similar triangles is equal to the square
of the ratio of their corresponding medians.
Solution. Given: AM and DN are the medians of triangles ABC and DEF
respectively
and, ΔABC ~ ΔDEF
area (ΔABC) AM 2
To Prove: =
area (ΔDEF) DN 2
Proof: ΔABC ~ ΔDEF (given)

area (ΔABC) ⎛ AB ⎞
2

Therefore, =⎜ ⎟ ...(i)
area (ΔDEF) ⎝ DE ⎠
AB BC CA
and, = = ...(ii)
DE EF FD
1
BC
AB CD
⇒ = 2 =
DE 1 FD
EF
2
In ΔABM and ΔDEN
∠B = ∠E [Since, ΔABC ~ ΔDEF]
AB BM
= [Prove in (i)]
DE EN
Therefore, ΔABC ~ ΔDEF [By SAS criterion of similarity]
AB AM
Therefore, = ...(iv)
DE DN
Therefore, ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the
corresponding sides.
area (ΔABC) AB2 AM 2
Therefore, = =
area (ΔDEF) DE 2 DN 2
Hence proved.
Q.7. Prove that the area of an equilateral triangle described on one side of a square
is equal to half the area of the equilateral triangle described on one of its
diagonals.

25
Solution. Given: ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are
two equilateral triangles described on the diagonals AC and side BC of
the square ABCD.
1
To Prove: area (ΔBQC) = area (ΔAPC)
2
Proof: ΔAPC and ΔBQC are both equilateral triangles. (Given)
Therefore, ΔAPC ~ ΔBQC [AAA similarity criterion]

area (ΔAPC) ⎛ AC ⎞
2
AC2
Therefore, =⎜ ⎟ =
area (ΔBQC) ⎝ BC ⎠ BC 2
2
⎛ 2BC ⎞ 2BC 2
⇒ ⎜ ⎟ = =2 [Since, Diagonal = 2 side = 2 BC]
⎜ BC ⎟ BC 2
⎝ ⎠
= area (ΔAPC) = 2 × area (ΔBQC)
1
or, area (ΔBQC) = area (ΔAPC)
2
Hence proved.
Q.8. Tick the correct answer and justify:
ABC and BDE are two equilateral triangles such that D is the mid-point of BC.
Ratio of the areas of triangles ABC and BDE is
(A) 2 : 1 (B) 1 : 2
(C) 4 : 1 (D) 1 : 4.
Solution. ΔABC and ΔBDE are two equilateral triangles. D is the mid-point of BC.
1
Therefore, BD = DC = BC
2
Let each side of triangle is 2a.
As, ΔABC ~ ΔBDE
area (ΔABC) AB2 (2a ) 2 4a 2 4
Therefore, = = = 2 = = 4 :1
area (ΔBDE) BD 2 (a) 2 a 1
Thus, correct option is (C).
Q.9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are
in the ratio
(A) 2 : 3 (B) 4 : 9
(C) 81 : 16 (D) 16 : 81

26
Solution. Let ABC and DEF are two similarly triangles ΔABC ~ ΔDEF (Given)
AB AC BC 4
and, = = = (Given)
DE DF EF 9
area (ΔABC) AB 2
Therefore, =
area (ΔDEF) DE 2
[The ratio of the areas of two triangles is equal
to square of the ratio of their
corresponding sides]
area (ΔABC) ⎛ 4 ⎞ 16
2

Therefore, =⎜ ⎟ = = 16 : 81
area (ΔDEF) ⎝ 9 ⎠ 81
Thus, correct option is (D).

Exercise 6.5
Q.1. Sides of triangles are given below. Determine which of them are right triangles.
In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm.
Solution. In a right triangle the square of the hypotenuse is equal to the sum of squares of
the other two sides.
(i) 7 cm, 24 cm, 25 cm
Let a = 7 cm, b = 24 cm, c = 25 cm
As, (7) 2 = 49; (24) 2 = 576; (25) 2 = 625
We have, a2 + b2 = (7)2 + (24)2 = 252 = c2
Hence, the given triangle is right angled.
Length of hypotenuse = 25 cm.
(ii) 3 cm, 8 cm, 6 cm
Let a = 3 cm, b = 6 cm, c = 8 cm
As, (3) 2 = 9; (8) 2 = 64; (6) 2 = 36
We have, a2 + b2 = 32 + 62 = 45 ≠ c2
82 ≠ 32 + 62 ⇒ 64 = 9 + 36 ⇒ 64 ≠ 45
Hence, the given triangle is not right angled.
(iii) 50 cm, 80 cm, 100 cm
Let c = 100 cm, b = 80 cm, a = 50 cm
Hence, the given triangle is not right angled.

27
(iv) 13 cm, 12 cm, 5 cm
Let c = 13 cm, a = 12 cm, b = 5 cm
132 = 169;122 = 144; 52 = 25
Here, c2 = a 2 + b2
We have, a2 + b2 = (80)2 + (50)2 = 6400 + 2500 = 8900 ≠ c2
⇒ (13) 2 = (5) 2 + (12) 2
⇒ 169 = 25 + 144 = 169
Hence, the given triangle is right angled.
Length of the hypotenuse = 13 cm
Q.2. PQR is a triangle right angled at P and M is a point on QR such that
PM ⊥ QR. Show that PM 2 = QM ⋅ MR.
Solution. Given: ΔPQR is right angled at P and M is a point on QR such that
PM ⊥ QR.
To Prove: PM 2 = QM ⋅ MR.
Proof: In ΔPQM, We have
PQ2 = PM2 + QM2 [By Pythagoras theorem]
2 2 2
Or, PM = PQ – QM ….(i)
In ΔPMR, We have
PR2 = PM2 + MR2 [By Pythagoras theorem]
2 2 2
Or, PM = PR – MR …(ii)
Adding (ii) and (ii), We get
2PM2 = (PQ2 + PM2) – (QM2 + MR2)
= QR2 – QM2 – MR2 [∴ QR2 = PQ2 + PR2]
= (QM + MR)2 – QM2 – MR2
= 2QM, MR
2
Therefore, PM = QM, MR,
Hence proved.
Q.3. In the given figure, ABD is a triangle right angled at A and AC ⊥ BD. Show
that
(i) AB 2 = BC ⋅ BD (ii) AC2 = BC ⋅ DC
(iii) AD2 = BD ⋅ CD
Solution. Given: A right angled ΔABD right angled at A and AC ⊥ BD.
To Prove:
(i) AB2 = BC ⋅ BD

28
 (ii) AC2 = BC ⋅ DC
(iii) AD2 = BD ⋅ CD
Proof: In ΔDAB and ΔDCA, We have
∠D = ∠D (Common)
∠BAD = ∠DCA = 90°
Therefore, ΔDAB ~ ΔDCA ...(i) [By AA similarity]
In ΔDAB and ΔACB,
∠B = ∠B (Common)
∠BAD = ∠BCA = 90°
Therefore, ΔDAB ~ ΔACB ...(ii)
From equation (i) and equation (ii), We have
ΔDAB ~ ΔACB ~ ΔDCA,
(i) ΔACB ~ ΔDAB (proved above)
area (ΔACB) AB2
Therefore, =
area (ΔDAB) DB2
1
BC × AC
2 AB2
⇒ = 2
1
DB × AC DB
2
BC AB 2
⇒ =
DB DB 2
BC AB 2
⇒ =
1 DB
AB2
⇒ BC =
BD
Therefore, AB2 = BC × BD
(ii) ΔACB ~ ΔDCA (proved above)
area (ΔACB) AC 2
Therefore, =
area (ΔDCA) DC 2
1
BC × AC
2 AC2
⇒ = 2
1
DC × AC DC
2
BC AC2
⇒ =
DC DC2

29
 AC2
⇒ BC =
DC
Therefore, AC 2 = BC × DC
(iii) ΔDAB ~ ΔDCA (proved above)
area (ΔDAB) DA 2
=
area (ΔDCA) DB 2
1
CD × AC
2 AD 2
⇒ = 2
1
BD × AC BD
2
CD AD 2
⇒ =
BD BD 2
AD 2
⇒ CD =
BD
Therefore, AD2 = BD × CD
Q.4. ABC is an isosceles triangle right angled at C. Prove that AB 2 = 2AC 2 .
Solution. Given: ΔABC is an isosceles triangle right angled at C.
To Prove: AB2 = 2AC2 .
Proof: In ΔACB, ∠C = 90°
AC = BC (Given)
AB = AC + BC
2 2 2
[By using Pythagoras theorem]
= AC2 + AC2 [Since, BC = AC]
AB = 2AC
2 2

Hence, proved.

Q.5. ABC is an isosceles triangle with AC = BC. If AB 2 = 2AC2 , prove that ABC is
right triangle.
Solution. Given: ABC is an isosceles triangle having AC = BC
and AB2 = 2AC2
To prove: ΔABC is a right triangle.
Proof: AB2 = 2AC2 (given)
⇒ AB2 = AC2 + AC2
Or, AB2 = AC2 + BC 2 [Since, AC = BC]
Hence, (By Pythagoras Theorem) ΔABC is right angled triangle.

30
Q.6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution. ABC is an equilateral triangle of side 2a .
Draw, AD ⊥ BC
In ΔADB and ΔADC, We have
AB = AC [Given]
AD = AD [Common]
∠ADB = ∠ADC [Each 90°]
Therefore, ΔADB ≅ ΔADC [By RHS congruence]
Hence, BD = DC = a [CPCT]
In right angled ΔADB
Here, AB 2 = AD 2 + BD 2
⇒ (2a) 2 = AD 2 + (a) 2
⇒ 4a 2 − a 2 = AD 2
⇒ AD 2 = 3a 2
⇒ AD = a 3
Q.7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum
of the squares of its diagonals.
Solution. Given: ABCD is a rhombus whose diagonals AC and BD intersect each other
at O.
To Prove: AB2 + BC 2 + CD 2 + AD2 = AC2 + BD 2
Proof: Since, the diagonals of a rhombus bisect each
other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB, ∠AOB = 90°
Therefore, AB2 = AO2 + BO 2 ...(i) [By Pythagoras Theorem]
Similarly, BC = CO + BO
2 2 2
...(ii)
CD = CO + OD
2 2 2
...(iii)
DA 2 = DO 2 + AO2 ...(iv)
Adding, equation (i), equation (ii), equation (iii) and equation (iv), we have
AB 2 + BC 2 + CD 2 + DA 2 = 2AO 2 + 2CO 2 + 2BO 2 + 2DO 2
= 4AO 2 + 4BO 2 [Since, AO = CO and BO = DO]
= (2AO) + (2BO) = AC + BD 2
2 2 2

31
Q.8. In the given figure, O is a point in the interior of a triangle
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA 2 + OB 2 + OC2 − OD2 − OE2 − OF 2 = AF 2 + BD2 + CE2
(ii) AF 2 + BD2 + CE2 = AE2 + CD2 + BF 2
Solution. Given: ΔABC in which
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
To Prove:
(i) AF2 + BD 2 + CE 2 = OA 2 + OB2 + OC 2 − OD 2 − OE 2 − OF2
(ii) AF2 + BD 2 + CE 2 = AE 2 + CD 2 + BF2

Proof: (i) In right angled ΔAFO, we have

OA 2 = OF 2 + AF 2 [By Pythagoras Theorem]
2
⇒ AF = OA − OF
2 2
...(i)
In right angle ΔBDO, we have,
OB 2 = BD 2 + OD 2 [By Pythagoras Theorem]
⇒ BD = OB − OD
2 2 2
...(ii)
In right angle ΔCEO, we have,
OC 2 = CE 2 + OE 2 [By Pythagoras Theorem]
⇒ CE = OC − OE
2 2 2
...(iii)
Adding equation (i), equation (ii) and equation (iii), we have
AF2 + BD2 + CE2 = OA 2 + OB 2 + OC 2 − OD 2 − OE 2 − OF 2
Which proves part (i),
Again, AF 2 + BD 2 + CE 2 = (OA 2 − OE 2 ) + (OC 2 − OD 2 ) + (OB 2 − OF 2 )
= AE 2 + CD 2 + BF 2
[AE 2 = AO 2 − OE 2 ; CD 2 = OC 2 − OD 2 ; BF 2 = OB 2 − OF 2 ]

32
Q.9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance
of the foot of the ladder from base of the wall.
Solution. Height of window from ground (AB) = 8 m .
Length of ladder (AC) = 10 m
Distance between foot of ladder and foot of wall (BC) = ?
In ΔABC, We have
AB2 + BC 2 = AC2 [By Pythagoras Theorem]
⇒ (8) 2 + (BC) 2 = (10) 2
⇒ 64 + BC2 = 100
⇒ BC2 = 100 − 64
⇒ BC = 36
⇒ BC = 6 m
Therefore, Distance between foot of ladder and base of the wall = 6 m.
Q.10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a
stake attached to the other end. How far from the base of the pole should the
stake be driven so that the wire will be taut?
Solution. Height of pole (AB) = 18 m
Length of wire (BC) = 24 m
In right angle triangle ABC, We have
AB2 + BC2 = AC2 [By Pythagoras Theorem]
⇒ 2 2
(18) + (BC) = (24) 2

⇒ 324 + (BC)2 = 576

⇒ BC2 = 576 – 324
⇒ BC = 252

⇒ BC = 6 7
Hence, the stake should be driven by 6 7 m from the base of the pole so that the
wire will be taut.
Q.11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per
hour. At the same time, another aeroplane leaves the same airport and flies due
west at a speed of 1200 km per hour. How far apart will be the two planes after
1
1 hours?
2
Solution. Speed of first aeroplane = 1000 km/hr.

33
 1 3
Distance covered by first aeroplane due north in 1 hours (OA) = 1000 × km
2 2
= 1500 km
Speed of second aeroplane = 1200 km/hr.
1
Distance covered by second aeroplane due west in 1 hours (OB)
2
3
= 1200 ×km
2
Or, OB = 1800 km
In right angle ΔAOB, We have
AB2 = AO2 + OB2
⇒ AB2 = (1500)2 + (1800)2
⇒ AB = 2250000 + 3240000
= 5490000
Or, AB = 300 61 km
Hence, distance between two aeroplanes will be 300 61 km
Q.12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance
between the feet of the poles is 12 m, find the distance between their tops.
Solution. Height of first pole (AB) = 11 m
Height of second pole (CD) = 6 m
Distance between foot of pole (BD) = 12 m
As, BE = DC = 6 m
Therefore, AE = AB − BE
= (11 – 6) m = 5 m
In right angle ΔAEC, We have
AC2 = AE 2 + EC2
AC = (5) 2 + (12) 2 = 25 + 144 = 169 = 13.
Hence, distance between their top = 13 m.
Q.13. D and E are points on the sides CA and CB respectively of a triangle ABC
right angled at C.
Prove that AE2 + BD2 = AB 2 + DE2 .
Solution. Given: ΔABC is right angled at C, D and E are the mid-points of CA and BC
respectively.

34
 1
i.e., CD = AD = AC
2
1
and, BE = EC = BC
2
To Prove: AE 2 + BD 2 = AB2 + DE 2
Proof: In right angles ΔBCA, We have
AB2 = BC 2 + CA 2 ...(i) [By Pythagoras Theorem]
In right angle ΔECD, We have
DE 2 = EC 2 + DC 2 ...(ii) [By Pythagoras Theorem]
In right angled ΔACE, We have
AE 2 = AC2 + CE 2 ...(iii) [By Pythagoras Theorem]
In right angled ΔBCD, We have
BD 2 = BC2 + CD 2 ...(iv) [By Pythagoras Theorem]
Adding equation (iii) and equation (iv), we have
AE 2 + BD 2 = AC 2 + CE 2 + BC 2 + CD 2
= [AC2 + CB 2 ] + [CE 2 + DC 2 ]
= AB2 + DE 2
Hence, AE 2 + BD 2 = AB2 + DE 2 .
Which is the required result.
Q.14. The perpendicular from A on side BC of a ΔABC intersects BC at D such
that DB = 3CD. (See figure). Pove that 2AB 2 = 2AC 2 + BC 2 .
Solution. Given: ΔABC, AD ⊥ BC, BD = 3CD.
To Prove: 2AB2 = 2AC2 + BC 2
Proof: In right angles triangles ADB and ADC, we have
AB2 = AD 2 + BD 2 ...(i)
AC = AD + DC
2 2 2
...(ii) [By Pythagoras Theorem]
Subtracting equation (ii) from equation (i), we get
AB2 − AC2 = BD 2 − DC 2
= 9CD 2 − CD 2 [∵ BD = 3CD]
2
⎛ BC ⎞
= 8CD 2 = 8 ⎜ ⎟
⎝ 4 ⎠
[Since, BC = DB + CD = 3CD + CD = 4CD]

35
 BC 2
Therefore, AB2 − AC2 =
2
⇒ 2(AB2 − AC 2 ) = BC 2
⇒ 2AB2 − 2AC2 = BC 2
∴ 2AB2 = 2AC2 + BC 2
Which is the required result.
1
Q.15. In an equilateral triangle ABC, D is a point on side BC such that BD = BC.
3
Prove that 9AD2 = 7AB 2 .
Solution. Given: Equilateral triangle ABC, D is a point on side BC such that
1
BD = BC.
3
To Prove: 9AD2 = 7AB2
Proof: In ΔABC, We have

1
BM = MC = BC [Since, ABC is an equilateral triangle and AM ⊥ BC] ... (i)
2
OM = BM – BD
1 1 1
= BC – BC [As, BD = BC]
2 3 3
BC
= … (ii)
6
Now, In ΔADM, We have
AD2 = AM2 + DM2 [By Pythagoras Theorem]
But, In ΔAMC, We have
AM2 = AC2 – MC2 [By Pythagoras Theorem]
2 2 2 2
Therefore, AD = AC + DM – MC
2 2
⎛ BC ⎞ ⎛ BC ⎞
= AC2 + ⎜ ⎟ –⎜ ⎟ [From (i) and (ii)]
⎝ 6 ⎠ ⎝ 2 ⎠
BC2 BC2
−
= AC2 + 36 4
AB2 AB2
−
= AB2 + 36 4 [Since, AB = BC = AC]
⎛ 1 1⎞
⎜1 + − ⎟ AB
2

= ⎝ 36 4 ⎠

36
 7
AB2
= a
⇒ 9AD2 = 7AB2,
Hence, proved.
Q.16. In an equilateral triangle, prove that three times the square of one side is
equal to four times the square of one of its altitudes.
Solution. Given: ABC is an equilateral triangle and AD ⊥ BC
To Prove: 3AB2 = 4AD2
Proof: In ΔABC,
Let AB = BC = AC = 2a
Also, AD ⊥ BC
1
Therefore, BD = DC = BC = a
2
In ΔABD , We have
AB2 = AD2 + BD2
⇒ (2a)2 = AD2 + (a)2
⇒ 4a2 = AD2 + a2
⇒ 4a2 – a2 = AD2
2
⎛ AB ⎞
AD 2 = 3a 2 = 3 ⎜ ⎟ [Since, AB = 2a]
⎝ 2 ⎠
AB2
⇒ AD2 = 3
4
⇒ 3AB = AD2
2

Which is the required result.

Q.17. Tick the correct answer and justify:
In ΔABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm. The angle of B is:
(A) 120° (B) 60°
(C) 90° (D) 45°
Solution. Here, AC = 12 cm
AB = 6 3 cm
and, BC = 6 cm
Now, AB2 + BC 2 = (6 3 ) 2 + (6) 2 = 108 + 36 = 144 = AC2
Hence by converse of pythagoras theorem ΔABC is right angled triangle right
angled at B.
Therefore, ∠B = 90°

37
 Exercise 6.6 (Optional)
Q.1. In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that
QS PQ
= .
SR PR
Solution. Given: ΔPQR, PS is the bisector of ∠QPR.
i.e., ∠1 = ∠2
QS PQ
To Prove: =
SR PR
Condition : Draw RT Parallel to PS. Entend
PQ to meet RT at T.
Proof: In ΔQRT, We have PS || TR
Therefore, ∠2 = ∠3 (Alternate angles)
Therefore, ∠1 = ∠4 (Corrresponding angles)
Therefore, ∠1 = ∠2 (Given)
Therefore, ∠3 = ∠4
In ΔPRT, We have
∠3 = ∠4
Therefore, PT = PR ⎡Side opposite to equal angles ⎤
As, PS || TR ⎢of a triangle are equal ⎥
⎣ ⎦
QP QS
Therefore, = [By Basic Proportionality Theorem]
PT SR
QP QS
⇒ = [Since, PT = PR]
PR SR
PQ QS
⇒ =
PR SR
Which is the required result.
Q.2. In the given figure, D is a point on hypotenuse AC of ΔABC, DM ⊥ BC ,
DN ⊥ AB, prove that:
(i) DM 2 = DN ⋅ MC (ii) DN 2 = DM ⋅ AN
Solution. Given: ΔABC, DM ⊥ BC and DN ⊥ AB
To Prove: (i) DM 2 = DN ⋅ AC
(ii) DN 2 = DM ⋅ AM
Proof: In ΔABC, BD ⊥ AC (Given)
⇒ ∠BDC = 90°

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 ⇒ ∠BDM + ∠MDC = 90° ...(i)
In ΔDMC, We have
∠DMC = 90°
⇒ ∠C + ∠MDC = 90° ...(ii)
From equation (i) and equation (ii), we have
∠BDM + ∠MDC = ∠C + ∠MDC
⇒ ∠BDM = ∠C
Now in ΔBMD and ΔMDC, We have
∠BDM = ∠C [Proved above]
∠BMD = ∠DMC = 90°
Therefore, ΔBMD ~ ΔMDC [By AA criterion of similarity]
DM MC
⇒ = [Since, corresponding sides of similar triangles are proportional]
BM DM
⇒ DM 2 = BM × MC
⇒ DM 2 = DN ⋅ MC [Since, BM = DN]
Similarly, in ΔNDA ~ ΔNBD , We have
DN AN
⇒ = [Since, corresponding sides of similar triangles are proportional]
BN DN
⇒ DN 2 = BN × AN
⇒ DN 2 = DM ⋅ AN
Hence proved
Q.3. In the given figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ BC
produced. Prove that AC2 = AB 2 + BC2 + 2BC ⋅ BD .
Solution. Given: ΔABC, AD ⊥ BC when produced, ∠ABC > 90°.
To Prove: AC2 = AB2 + BC 2 + 2BC ⋅ BD
Proof: In ΔADC , We have
(AC)2 = (AD)2 + (DC) 2 ...(i)
⇒ (AC) 2 = (AD)2 + (BC + BD) 2
⇒ AC2 = AD 2 + BC 2 + BD 2 + 2BC ⋅ BD
= AB2 + BC2 + 2BC ⋅ BD (Since, AB2 = AD 2 + BD 2 )
Hence Proved.
Q.4. In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC.
prove that
AC2 = AB 2 + BC2 − 2BC ⋅ BD

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Solution. Given: ΔABC, ∠ABC < 90° and AD ⊥ BC.
To Prove: AC2 = AB2 + BC 2 − 2BC ⋅ BD
Proof: ΔADC is right-angled at D
Therefore, AC2 = CD 2 + DA 2 ...(i)
[By Pythagoras Theorem]
Also, ΔADB is right angled at D
Therefore, AB2 = AD 2 + DB2 [By Pythagoras Theorem] ...(ii)
From equation (ii), We have
AD 2 = AB2 − DB2
⇒ AC2 = CD 2 + (AB2 − BD 2 ) [From equation (i)
⇒ AC2 = (BC − BD) 2 + AB2 − BD 2 (Since, BC = BD + DC)
⇒ (AC) 2 = BC2 + BD 2 − 2BC ⋅ BD + AB2 − BD 2
⇒ AC2 = AB2 + BC 2 − 2BC ⋅ BD , Hence Proved.
Q.5. In the given Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove
that:
2
2 2 ⎛ BC ⎞
(i) AC = AD + BC. DM + ⎜ ⎟
⎝ 2 ⎠
2
2 2 ⎛ BC ⎞
(ii) AB = AD − BC. DM + ⎜ ⎟
⎝ 2 ⎠
1
(iii) AC2 + AB 2 = 2AD2 + BC2
2
Solution. (i) In ΔAMC, We have
(AC) 2 = (AM) 2 + (MC) 2 [By Pythagoras Theorem]
⇒ (AC) 2 = (AM) 2 + (MD + DC) 2 [Where, MC = MD + DC]
⇒ (AC) 2 = (AM) 2 + (MD) 2 + (DC) 2 + 2MD ⋅ DC
In ΔAMD, AD 2 = AM 2 + MD 2
2
⎡ In ΔAMD, AD2 = AM 2 + MD 2 ⎤
⎛ BC ⎞ ⎢ ⎥
⇒ (AC) 2 = (AD) 2 + ⎜ ⎟ + 2MD ⋅ DC ⎢ 1 ⎥
⎝ 2 ⎠ ⎢⎣
and DC = BC
⎥⎦
2
2
⎛ BC ⎞
⇒ (AC) 2 = AD 2 + ⎜ ⎟ + BC ⋅ DM Proved [Since, DC = BD]
⎝ 2 ⎠
(ii) In ΔAMB, We have

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 AB2 = AM 2 + MB 2 [By Pythagoras Theorem]
⇒ AB2 = AM 2 + (BD − MD) 2
BC
AB2 = AM 2 + (BD) 2 + (MD) 2 − 2BD. MD [Since, AD = AM + MD and BD =
2 2 2
⇒ ]
2
= AM 2 + MD 2 + BD 2 − 2BD ⋅ MD
2
⎛ BC ⎞ BC
⇒ AB2 = AD 2 + ⎜ ⎟ − 2× ⋅ MD
⎝ 2 ⎠ 2
2
⎛ BC ⎞
⇒ AB = AD + ⎜
2 2
⎟ − BC ⋅ MD
⎝ 2 ⎠
2
⎛ BC ⎞
⇒ AB = AD − BC ⋅ MD + ⎜
2 2
⎟ , Hence Proved
⎝ 2 ⎠
(iii) Adding the result obtained in (i) and (ii), We get

2
⎛ BC ⎞
AC + AB = 2(AD) + 2 ⎜
2 2 2
⎟
⎝ 2 ⎠
BC 2
⇒ AC 2 + AB2 = 2AD 2 + 2 ×
4
BC 2
⇒ AC 2 + AB2 = 2AD 2 +
2
Hence Proved.
Q.6. Prove that the sum of the squares of the diagonals of a parallelogram is equal
to the sum of squares of the sides.
Solution. Given: ABCD is a parallelogram whose diagonals are AC and BD.
To Prove: (AB)2 + (BC) 2 + (CD) 2 + (DA)2 = (AC) 2 + (BD) 2
Proof : In ΔAMD, We have
AD = BC [Opposite sides of a parallelogram]
AM = BN [Both are altitudes of the same parallelogram to the same base]
Therefore, ΔAMD ≅ ΔBNC [By RHS congruence criterion]
Construction : Draw AM ⊥ CD and BN ⊥ CD.
Hence, MD = NC ...(i) [CPCT]
In ΔBND, We have
(BD) 2 = (BN) 2 + (DN) 2
[By Pythagoras Theorem]

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 = BN 2 + (DC + CN) 2
= BN 2 + DC2 + CN 2 + 2DC ⋅ CN
= BN 2 + CN 2 + DC2 + 2DC ⋅ CN
Or, (BD) 2 = (BC)2 + (DC) 2 + 2DC ⋅ CN ...(ii) [Since, (BC) 2 = (BN) 2 + (CN) 2 ]
In ΔAMC, We have
(AC)2 = AM 2 + MC2
= AM 2 + (DC − DM) 2
= AM 2 + DC2 + DM 2 − 2DC ⋅ DM
= AM 2 + DM 2 + DC2 − 2DC ⋅ DM
Or, (AC)2 = AD2 + DC2 − 2DC ⋅ DM
[Where, AD 2 = AM 2 + DM 2 ]
Or, AC2 = AD 2 + AB2 − 2DC ⋅ CN ...(iii) [Since, AB = DC]
Adding equation (ii) and equation (iii), we have
AC 2 + BD 2 = (AD 2 + AB2 ) + (BC 2 + DC 2 )
= AB2 + BC 2 + CD 2 + DA 2
⇒ (AC) 2 + (BD) 2 = (AB)2 + (BC) 2 + (CD) 2 + (DA)2
Hence Proved.
Q.7. In the given figure, two chords AB and CD intersect each
other at the point P.
Prove that :
(i) ΔAPC ~ ΔDPB (ii) AP ⋅ PB = CP ⋅ DP
Solution. Given: AB and CD are two chords of a circle intersects each other at P.
To Prove: (i) ΔAPC ~ ΔDPB
(ii) AP ⋅ PB = CP ⋅ DP
Proof: (i) In ΔAPC and ΔDPB, We have
∠APC = ∠DPB (Vertically opposite angle)
∠CDB = ∠CAB (Angle in the same segment)
Therefore, ΔAPC ~ ΔDPB (By AA similarity criterion)
(ii) As, ΔAPC ~ ΔDPB [Proved in part (i)]
AP PC
=
DP PB ( If two triangles are similar corresponding sides are proportional )
Therefore,

⇒ AP ⋅ PB = PC ⋅ DP
Hence Proved.

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Q.8. In the given figure, two chords AB and CD of a circle intersect each other at
point P (when produced) outside the circle. Prove that:
(i) ΔPAC ~ ΔPDB
(ii) PA ⋅ PB = PC ⋅ PD

Solution. Given: AB and CD are two chord of circle intersects each other at P (when
produced) out side the circle.
To Prove: (i) ΔPAC ~ ΔPDB
(ii) PA ⋅ PB = PC ⋅ PD
Proof: (i) In ΔPAC and ΔPDB , We have
∠PCA = ∠PBD
Exterior angle of a cyclic quadrilateral is equal to interior opposite angle)
Also, ∠P = ∠P [Common]
Therefore, ΔPAC ~ ΔPDB (AA similarity criterion)
(ii) As, ΔPAB ~ ΔPDB

Therefore,
PA PC
= [ We know if two triangle are similar corresponding sides are proportional]
PD PB
⇒ PA × PB = PC × PD
PA ⋅ PB = PC ⋅ PD
Q.9. In the given figure, D is a point on side BC of ΔABC
BD AB
such that = .
CD AC
Prove that AD is the bisector of ∠BAC.

Solution. Given: ΔABC, D is a point on BC such that

BD AB
=
DC AC
To Prove: AD is bisector of ∠BAC, i.e.,
∠BAD = ∠CAD
Construction : Draw CE || AD to meet AB, (when extended at E)
Proof: In ΔBCE, we have

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 AD || CE ...(By construction)
So, By Basic Proportionality Theorem, We have
AB AB
BD AB ⎫ ⇒ = ⎡ BD AB ⎤
= ⎬ AE AC ⎢Since, =
DC AE ⎭ ⎣ CD AC ⎥⎦
⇒ AE = AC
In ΔACE, we have:
AE = AC [Proved, above]
⇒ ∠AEC = ∠ACE ...(i)
Also, ∠CAD = ∠ACE [Since, AD || CE] ...(ii)
Similarly,
∠BAD = ∠AEC [Since, AD || CE] ...(iii)
Using equation (i), equation (ii) and equation (iii), we have
∠BAD = ∠CAD
Hence Proved.
Q.10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the
surface of the water and the fly at the end of the string rests on the water 3.6
m away and 2.4 m from a point directly
under the tip of the rod. Assuming that her
string (from the tip of her rod to the fly) is
taut, how much string does she have out
(see figure)? If she pulls in the string at the
rate of 5 cm per second, what will the horizontal
distance of the fly from her after 12 seconds?
Solution. In triangle ABC, We have
AB = 1.8 m
BC = 2.4 m
∠B = 90°
Therefore, AC2 = AB2 + BC2 (By Pythagoras Theorem)
⇒ AC = (1.8) + (2.4)
2 2 2

⇒ AC2 = 3.24 + 5.76 = 9

⇒ AC2 = (3) 2
⇒ AC = 3 m
Length of the string pulled in 12 second
= 5 × 12 = 60 cm = 0.6 m
Length of remaining string left out.
AD = (3 − 0.6) m

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 AD = 2.4 m
In ΔABD, We have
AD 2 = AB2 + BD 2 [By Pythagoras Theorem]
⇒ BD = AD − AB
2 2 2

⇒ (BD) 2 = (2.4) 2 − (1.8) 2

⇒ BD 2 = 5.76 − 3.24
⇒ BD 2 = 2.52 m
⇒ BD = 1.587 m
Therefore, Horizontal distance of the fly from Nazima
= BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m
Hence, original length of string and horizontal distance of the fly from Nazima is
3 m and 2.79 m respectively.

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