Self Study1 M.
Prakash Academy
Theme: Concurrency.
Ceva’s Theorem
Consider △ABC.
Let P be a point in the plane of △ABC such that
←
−−→ ← −−→ ←
−−→ ← −−→ ←
−−→ ← −−→
AP ∩ B C = {D}, B P ∩ C A = {E} and C P ∩ AB = {F }.
BD CE AF
Then · · = 1.
DC EA F B
Remark: Directed distances must be used in the above expression.
Tools Required:
(T1) Area of a triangle equals half base into height.
x1 y1 x1 y1 x1 − y1
(T2) If = then = = .
x2 y2 x2 y2 x2 − y2
F E
P
B D C
Fig.1
Strategy:
(1) If two triangles have equal heights then the ratio of their areas
equals the ratio of their bases.
BD
(2) Express as the ratio of areas of two triangles.
DC
Do this in two different ways.
BD [ABP ]
Use T 2 to prove that = .
DC [AP C]
CE AF
(3) Exploit the cyclic symmetry of △ABC to find and .
EA FB
BD CE AF
(4) Compute Ceva Product · · . Confirm that it is 1.
DC EA F B
Hurray! You have proved Ceva’s Theorem on your own.
Cogratulations.
For the sake of completeness we are giving the details of the above
computations.
Compare these calculations with your own computations.
BD [ABD] [P BD] [ABD] − [P BD] [ABP ]
= = = = (1)
DC [ADC] [P DC] [ADC] − [P DC] [AP C]
CE [BCE] [P CE] [BCE] − [P CE] [BCP ]
= = = = (2)
EA [BEA] [P EA] [BEA] − [P EA] [BP A]
AF [CAF ] [P AF ] [CAF ] − [P AF ] [CAP ]
= = = = (3)
FB [CF B] [P F B] [CF B] − [P F B] [CP B]
Multiplying (1), (2) and (3) give us,
BD CE AF [ABP ] [BCP ] [CAP ]
· · = · · = 1.
DC EA F B [AP C] [P BA] [CP B]
Remark: In the above figure we have taken point P in the interior
of △ABC. We are allowed to take point P any where in the plane
←
−−→ ←
−−→
of △ so for as AP is not parallel to B C etc. In the next figure
we have chosen point P in the exterior of △ABC.
B D C
P E
F Fig. 2
Convention: Consider [ABD].
Move along the perimeter of △ABD from A to B, then from B to
D and finally from D to A.
If one moves in anticlockwise sense then the area [ABD] is posi-
tive.
If one moves in the clockwise sense the area [ABD] is negative.
Observations
(O1) In both the figures BD
DC
is positive.
(O2) In both the figures [ABD] and [ADC] is positive.
(O3) In Fig1 [P BD] is positive and in Fig2 [P BD] is negative.
(O4) In Fig1 [P DC] is positive and in Fig2 [P DC] is negative.
[P BD]
(O5) In both the figures [P DC]
is positive.
(O6) Equation (1) is completely applicable in both the figures.
(O7) In equation (2) in Fig 2 CE and EA have opposite siqns.
(O8) In Fig2 [BCE] and [BEA] have opposite siqns.
(O9) In Fig2 [P CE] and [P EA] have opposite siqns.
(O10) Equation (2) is correct for Fig2 also.
(O11) Similarly equation (3) is correct for Fig2 also.
(O12) Our proof of Ceva’s Theorem is true in the case of Fig2 also.
Is the converse of Ceva’s Theorem also true?
We invite you to formulate the converse statement on your own.
Now carefully check, your statement with the one given below.
Converse of Ceva’s Theorem:
←
−−→ ← −−→ ← −−→
Consider △ABC. Let D, E, F be points on lines B C , C A , AB
BD CE AF
respectively such that the product · · = 1.
DC EA F B
←−−→ ← −−→ ←−−→
Then the lines AD , B E and C F are concurrent.
How do we go about proving the converse?
←−−→ ←−−→
Strategy: Let lines B E and C F meet at P.
←−−→ ← −−→
Let line AP intersect B C at D1 .
It is enough to show that D ≡ D1 .
← −−−→ ← −−→ ←
−−→
Tools: The three lines AD1 , B E and C F are concurrent at P,
hence we can use the Ceva’s Theorem.
We expect that, you complete the proof on your own. Having done
so, you can compare your proof with the one given below.
BD CE AF
Proof: It is given that · · = 1. (1)
DC EA F B
←
−−−→ ← −−→
By applying Ceva’s Theorem to concurrent lines AD1 , B E and
←−−→ BD1 CE ˙ AF
C F we obtain · =1 (2)
D1 C EA F B
(1) and (2) give
BD BD1 BD BD1 BD + DC BD1 + D1 C
= ∴ +1 = +1. ∴ =
DC D1 C DC D1 C DC D1 C
BC BC
∴ = ∴ DC = D1 C ∴ DC = −CD1
DC D1 C
∴ DC + CD1 = 0
∴ DD1 = 0
∴ D ≡ D1 . This completes the proof of the converse.
Convention: Consider △ABC. Let P be a point in the plane of
←−−→ ←−−→
△ such that AP intersects B C in point D.
←−−→
In the honour of Ceva, the line AD is called a cevian. (similar to
median)
The triangle whose vertices are three points D, E, F in the context
of Ceva’s Theorem is called the Cevian Triangle generated by point
P.
We encapsulate the above two theorems as Cevians are concurrent
iff the corresponding Ceva product equals unity.
Applications:
Converse of Ceva’s Theorem is a very powerful tool if we want to
prove that three cevians are concurrent. All one needs to do is
compute the ceva product. And generally it is enough to compute
one ratio BD
DC
. The other two can be computed by symmetry con-
siderations of △ABC.
A1: Medians of a triangle are concurrent.
A2: Internal angle bisectors of a triangle are concurrent.
A3: Two external bisectors and one inernal bisector of a triangle
are concurrent.
A4: Altitudes of a triangle are concurrent.
A5: Let the incircle touch BC, CA, AB at D, E, F respectively.
←
−−→ ← −−→ ← −−→
Then AD , B E , C F are concurrent.
The point of their intersection is called the Gergonne Point.
A6: Let excircles with centers I1 , I2 , I3 touch BC, CA, AB at D1 , E2 , F3
respectively.
←
−−→ ← −−−→ ← −−−→
Then AD 1 , B E2 , C F3 are concurrent.
This point of concurrency is called Nagel Point.
A7: In the context of incenter-excenter configuraion
←−−−→ ← −−−→ ← −−−→
(1) AD1 , B E1 , C F1 are concurrent.
←−−−→ ← −−−→ ← −−−→
(2) AD1 , B E3 , C F2 are concurrent.
Discover several concurrencies in this configuration on your own.
Write detailed proofs of A1 to A7 on your own.
