9

Mathematics
Quarter 3 – Module 2:
Midline Theorem, Theorems on
Trapezoids and Kites, and
Solving Problems Involving
Parallelograms, Trapezoids and
Kites
Mathematics – Grade 9
Quarter 3 – Module 2: Midline Theorem, Theorems on Trapezoids and Kites, and
Solving Problems Involving Parallegorams, Trapezoids, and Kites First Edition, 2020

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 9
Mathematics
Quarter 3 – Module 2:
Midline Theorem, Theorems on
Trapezoids and Kites, and
Solving Problems Involving
Parallelograms, Trapezoids and
Kites

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Introductory Message
For the facilitator:
As a facilitator, you are expected to orient the learners on how to use this
module. You also need to keep track of the learners' progress while allowing them
to manage their own learning at home. Furthermore, you are expected to encourage
and assist the learners as they do the tasks included in the module.

For the learner:

As a learner, you must learn to become responsible of your own learning. Take
time to read, understand, and perform the different activities in the module.
As you go through the different activities of this module be reminded of the
following:
1. Use the module with care. Do not put unnecessary mark/s on any part of the
module. Use a separate sheet of paper in answering the exercises.
2. Don’t forget to answer Let Us Try before moving on to the other activities.
3. Read the instructions carefully before doing each task.
4. Observe honesty and integrity in doing the tasks and checking your answers.
5. Finish the task at hand before proceeding to the next.
6. Return this module to your teacher/facilitator once you are done.
If you encounter any difficulty in answering the tasks in this module, do not
hesitate to consult your teacher or facilitator. Always bear in mind that you are not
alone. We hope that through this material, you will experience meaningful learning
and gain deep understanding of the relevant competencies. You can do it!

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 Let Us Learn
After going through this module, the learners are expected to be able
to:

1. Prove the Midline Theorem - M9GE-IIId-1

2. Prove theorems on Trapezoids and Kites - M9GE-IIId-2
3. Solve problems involving Parallelograms, Trapezoids and Kites - M9GE-IIIe-1

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 Let Us Study
In this section, you will learn different theorems on trapezoids and
kites and will apply these theorems to solve problems involving
trapezoids and kites.

Lesson
Midline Theorem for
1 Triangles

Have you ever encountered a situation where you have to cut an irregular
shape of manila paper into two equal parts for your paper mâché project in your
elementary days? How did you cut it? Finding a symmetrical “center line” would have
been your decision to cut the paper in half. There are several circumstances where
you would have encountered that you have to divide a shape (like a parallelogram)
into two parts.

In today's lesson, we will prove the Triangle Midsegment Theorem. In a

triangle, a midline (or a midsegment) is any of the three lines joining the midpoints of
any pair of the sides of the triangle.

In a triangle, the midline joining the midpoints of two sides is

parallel to the third side and half as long. Conversely, the line joining
points on two sides of a triangle, parallel to its third side and half as
long is a midline.

Given:
%𝐷%%𝐸% is a midsegment to ΔABC.

Proof:

In order to prove the theorem, we need to validate two statements. We have to prove that:

1
1. DE is one half of BC.

2
DE = BC

2. DE is parallel to BC.

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 To show that line segment lengths are equal, we typically use triangle
congruency, so we will need to construct a couple of triangles here.

To show the congruency, we will need to rely on what is given in the

problem statement and see which parts (sides or angles) we know are equal. In
this case, since DE is the midsegment, we know that AE=EC and AD=DB, so
we should use one or more of these sides in the triangles we construct.
We also need to prove something about DE - that is equal to half of BC -
so it would be useful for the triangle we construct to have DE as a side, too,
with another side equal to it - which we could construct.

With that in mind, let's construct a new triangle by extending line DE to point F,
so that DE=EF, and connecting F with C:

Now it is easy to show the new triangle we constructed (ΔCFE) is

congruent to ΔADE using the Side-Angle-Side (SAS) congruency postulate, and
as a result, that DFCB is a parallelogram since we now have two sides, DB and
FC that are both equal and parallel. And because DFCB is a parallelogram, DE
is parallel to BC, as we needed to prove. From the congruency of ΔCFE
and ΔADE, we also have DE=EF, and since DF= BC as opposite sides in a
parallelogram, DE is half of BC.

Pr o o f o f T h e T r i a n g le M id se g m e n t T h e o r em
STATEMENT REASON
1. DE=EF Construction

3. ∠AED ≅ ∠CEF
2. AE=EC Given, DE is a Midsegment

4. ΔADE ≅ΔCFE
Vertical Angles
Side-Angle-Side Congruence Postulate
5. ∠DAE ≅ ∠FCE Corresponding Angles in Congruent Triangles,
(CPCTC)
6. AB||CF Converse Alternate Interior Angles Theorem
7. AD=CF Corresponding Sides in Congruent Triangles,
(CPCTC)
8. AD=DB Given, DE Is a Midsegment
9. CF=DB Transitive Property of Equality
10. DFCB is a Two Opposite Sides of a Parallelogram, CF & DB,
parallelogram Are Equal and Parallel
11. DE||BC Opposite Sides of a Parallelogram
12. DF=BC Opposite Sides of a Parallelogram

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 13. DE=EF Corresponding Sides in Congruent Triangles,
(CPCTC)
14. DE= ½DF As Derived
15. DE= ½BC Substitution

Example 1:

Given M, N midpoints. MN = 12
Required: Find DF.

Solution:

Example 2:

Given D, E midpoints.
DE = 3x = 5; AB = 26
Required: Find x.

Solution:

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 Lesson
Midline Theorem for
2 Trapezoids

Similarly, in a trapezoid, a midline (or a midsegment) is the line joining the

midpoints of the sides.

In a trapezoid, the midline is parallel to the bases and its length is

half their sum. Conversely, the line joining points on the two sides of
a trapezoid, parallel to its bases and half as long of their sum is the
midline.

Proof:
We are going to prove the Midline Theorem of a Trapezoid.

In this part, we need to prove that

1. EF is half of the sum of AB and CD.
EF = ½ (AB + CD)
2. EF is parallel to both segments AB and CD.

Since we are dealing with the midpoints of segments, we will use what
we have already proven for triangle midsegments. Let's create such triangles,
by drawing a line from the vertex A through the midpoint, F, until it intersects
an extension of the base DC at point G:

We can easily show that ΔABF and ΔGCF are congruent, using the Angle-
Side-Angle postulate. From this, we can show that EF is a midsegment of

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triangle ΔADG. As such, by the Triangle midsegment theorem, it is parallel
to DG and is equal to half of DG.

PROOF OF MIDSEGMENT THEOREM OF TRAPEZOIDS

STATEMENT REASON

2. ∠BAF ≅ ∠CGF
1. AB||DG Given
Alternate Interior Angles Theorem

4. ΔABF ≅ ΔGCF
3. BF=FC Given
ASA Postulate
5. AF=FG Corresponding sides of congruent triangles
6. EF is midsegment Definition of midsegment
7. EF||DG Triangle midsegment theorem
8. EF=½DG Triangle midsegment theorem
9. DG = DC+CG Given
10. CG=AB Corresponding sides of congruent triangles
11. EF=½(DC+CG) Transitive property of equality
12. EF=½(DC+AB) Transitive property of equality

But DG is DC+CG, and as ΔABF and ΔGCF are congruent, CG=AB,

so EF is equal to half of DC+AB. In other words, the length of EF is
the arithmetic mean (average) of the lengths of the bases.

Example 3: Suppose that 𝐴𝐵=128 and 𝑋𝑌=95.What is 𝐶𝐷?

Given: 𝐴𝐵 = 128 and 𝑋𝑌 = 95.

Required: 𝐶𝐷

Solution:
1
2
Formula: XY = (CD + AB)

Substituting values to the formula,

1
2
95 = (CD + 128)

190 = CD + 128 multiplying both sides by 2 to eliminate ½

CD = 190 – 128

CD = 62

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Example 4: If 𝐵𝐴 = 24 cm and 𝐷𝐶 = 26 cm., find the length of 𝐾𝐹.

Given: 𝐵𝐴 = 24 cm; 𝐷𝐶 = 26 cm

Required: KF

Solution:

Formula: KF = 1 (BA + CD)

2

Substituting values to the formula,

1
2
KF = (24 cm + 26 cm)

1
2
= (50 cm)

= 25 cm

Lesson
Theorems of Trapezoids
3 and Kites

In this section, we will be discussing the properties and create theorems for
parallelograms specifically Trapezoids and Kites.

TRAPEZOIDS

 A trapezoid is a quadrilateral with only one pair of parallel sides.

%𝑨%%𝑩%|| %𝑪%𝑫%%

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 The median of a trapezoid is parallel to the bases and has a length equal to
half the sum of the lengths of the bases.

median

𝟐
IJ = (HG + EF)

 An isosceles trapezoid is a trapezoid with congruent legs.

AD = BC

 The diagonals of an isosceles trapezoid are congruent.

BD = AC

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  Opposite angles of an isosceles trapezoid are supplementary.

∠ A + ∠ C = 180°
∠ B + ∠ D = 180°

Example 5: The diagonals of an isosceles trapezoid are represented by 4x-47 and

2x + 31. What is the value of x?

Solution: The diagonals of an isosceles trapezoid are congruent. Therefore,

4x – 47 = 2x + 31 (since diagonals are equal)

4x -2x = 31+47 (combining like terms)

2x = 78 (using algebraic addition)

x = 39 (answer)

Example 6: Trapezoid FINE is an isosceles trapezoid with legs FI and EN. If FI = (7x
+ 14) cm and EN = (5x + 34) cm. Find the length of the legs of the trapezoid.

Solution: An isosceles trapezoid is a trapezoid with congruent legs Therefore,

FI = EN

7x + 14 = 5x + 34 (since both legs of an isosceles trapezoid are equal)

7x – 5x = 34 - 14 (combining like terms)

2x = 20 (using algebraic addition)

x = 10 cm

However, 10 cm is not the length of one leg. We need to integrate this value into each
equation in order to find the “true” value of each leg. Therefore,

FI = [7(10) + 14] cm (substituting the value of x in the equation of leg FI)

[70 + 14] cm

= 84 cm

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Since, leg FI is equal to leg EN, then EN is also 84 cm. Thus, each leg measures 84
cm

leg FI = leg EN = 84 cm. (answer)

LESSON 4: PROPERTIES OF KITES

 A kite is a quadrilateral with two distinct pairs of consecutive sides that are
congruent.

KL = LM
KJ = JM

 The diagonals of a kite are perpendicular.

%𝑲 %𝑱
% %𝑳

 A kite has exactly one pair of opposite angles that are congruent.

∠JKL = ∠JML

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  One diagonal (segment JL, the main diagonal) is the perpendicular bisector of
the other diagonal (segment KM, the cross diagonal). (The terms “main
diagonal” and “cross diagonal” are made up for this example.)

𝑱%%𝑳% bisects %𝑲%%𝑴%%

 The area of a kite is half the product of the lengths of the diagonals.

𝟐
Area = (KM ● JL)

Example 7: Quadrilateral ABCD is a kite with diagonals BD and AC.

If AD = 20 cm and ∠ADC = 50°. Find the value of w, x, y, z, and w.

∠
A C

∠ y
20 cm

Solution:

a. Since quadrilaterals have equal opposite angles, we can see that ∠ABC =
∠ADC and ∠w is half of ∠ADC. Therefore,
𝟓𝟎°
∠w = = 25° (∠w is one half of ∠ADC)
𝟐

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 x = 90° (since diagonals of a kite are perpendicular)
y = 20 cm (since two distinct pairs of consecutive sides are
congruent)
b. Perimeter = 2z + 2y (using the formula for Perimeter)

56 = 2z + 2(20) (substituting the value of y into the equation)

56 = 2z + 40
2z = 56 – 40 (combining like terms and solving for
z) 2z = 16
z = 8 cm (answer)

Example 8: Given RSTV is a kite. Find the indicated measure.

a. IF RS = 8 and ST = 12 then RV = .
VT = and the perimeter = .
Answer:
a.1. RV = 8 (RS = RV)

a.2. VT = 12 (VT = ST)

a.3. Perimeter

P = VT + ST + RS + RV (using the formula for Perimeter)

= 12 + 12 + 8 + 8 (substituting all values)
Perimeter = 40 (answer)

b. m∠SXT =
Answer:

m∠SXT = 90° (diagonals of a kite are perpendicular)

c. If m∠STX = 30° and m∠SRV = 80° then m∠VTX = , m∠RST = and

m∠SRX = .

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 Answer:
c.1. Solution: A diagonal of a kite bisect each of the noncongruent
angles and the other diagonal. Thus,

m∠VTX = 30° (∠VTX = ∠STX)

c.2. Solution: A kite has exactly one pair of opposite angles that are
congruent. Thus,

quadrilateral is 360°)
80° + 60° + m∠RST + m∠RVT = 360° (sum of all angles of a

80° + 60° + x° + x° = 360° (replacing unknown values with x)

140° + 2x° = 360° (combining like terms)

2x° = 360° - 140° (using algebraic addition)

2x° = 220°
𝟐𝟐𝟎°

𝟐
x=

x = 110°

m∠RST = m∠RVT = 110° (answer)

the other diagonal, therefore, m∠SRX is half of ∠SRV. Given that

c.3. A diagonal of a kite bisect each of the noncongruent angles and

∠SRV = 80°,
𝟖𝟎°
𝟐
m∠SRX = = 40° (answer)

Let Us Practice

Let Us Remember
In a triangle, the midline joining the midpoints of two sides is parallel to the third
side and half as long. Conversely the line joining points on two sides of a triangle,
parallel to its third side and half as long is a midline. And it has 3 properties, they
are;

1. The Midline or Midsegment of a Triangle is one-half of the base.

2. The Midline or Midsegment of a Trapezoid is one-half of the sum of the two
bases.
3. The Midline or Midsegment for both Triangle and Trapezoid is always parallel
to its base or bases.

The theorems of a trapezoid are:

A. A trapezoid is a quadrilateral with only one pair/s of parallel sides.

B. The median of a trapezoid is parallel to the bases and has a length equal
to half the sum of the lengths of the bases.
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 C. The diagonals of an isosceles trapezoid are congruent.
D. Opposite angles of an isosceles trapezoid are supplementary.
E. An isosceles trapezoid is a trapezoid with congruent legs.

The theorems of a Kite are:

A. A kite is a quadrilateral with two distinct pairs of consecutive sides that
are congruent.
B. The diagonals of a kite are perpendicular.
C. A kite has exactly one pair of opposite angles that are congruent.
D. A diagonal of a kite bisects each of the noncongruent angles and the
other diagonal.
E. The area of a kite is half the product of the lengths of the diagonals.

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