1. GRADE 9 MATHEMATICS
QUARTER 3
Module 5
May 2014

2. FIGURE, PICTURE,
QUOTATION ‘n ONE
Instruction:
Arrange the jigsaw puzzle to form a picture/figure and
paste it in the manila paper. Shout your yell after you
made this activity.
Present the puzzle, describe the figure and explain
briefly the Mathematics quotation found therein.

3. Competencies
1. Identifies quadrilaterals that are parallelograms.
2. Determines the conditions that guarantee a
quadrilateral a parallelogram.
3. Uses properties to find measures of angles, sides and
other quantities involving parallelograms.
4. Proves theorems on the different kinds of
parallelograms (rectangle, rhombus, square)
5. Proves the Midline Theorem.
6. Proves theorems on trapezoids and kites.
7. Solves problems involving parallelograms, trapezoids
and kites.

4. Parallelograms
Definition:
A quadrilateral whose opposite sides are parallel.
(Garces, et. Al., 2007)

5. Properties of Parallelograms
1. The diagonal divides it into two congruent triangles.
2. Opposite angles are congruent.
3. Two consecutive angles are supplementary.
4. The diagonals bisect each other.
5. Opposite sides are congruent.
6. The opposites sides are parallel and congruent.

6. Parallelograms
The diagonal divides it into
two congruent triangles.
Proof:
Given that □ABCD is a parallelogram. We have to prove
that ∆ABD ≅ ∆CDB.
By definition , 𝐴𝐵 ∥ 𝐶𝐷 and 𝐵𝐶 ∥ 𝐴𝐷, then we can say that
∠𝐷𝐵𝐶 ≅ ∠𝐵𝐷𝐴 and ∠𝐶𝐷𝐵 ≅ ∠𝐴𝐵𝐷 by Alternate Interior
Angles theorem. This implies that ∆ABD ≅ ∆CDB by ASA
congruence theorem.
B
A D
C

7. Parallelograms
Opposite angles are congruent.
Proof:
Given that □ABCD is a parallelogram. We have to prove
that ∠𝐵𝐴𝐷 ≅ ∠𝐷𝐶𝐵.
Because ∆ABD ≅ ∆CDB by ASA congruence theorem, then
∠𝐵𝐴𝐷 ≅ ∠𝐷𝐶𝐵 by CPCTC.
B
A D
C

8. Parallelograms
Consecutive angles are supplementary.
Proof:
Given that □ABCD is a parallelogram. We have to prove
that ∠𝐴𝐵𝐶 + ∠𝐵𝐶𝐷 = 180.
Since the sum of interior angles of quadrilaterals is 3600
and the opposite angles of parallelogram are congruent,
then we can say that 2∠𝐴𝐵𝐶 + 2∠𝐵𝐶𝐷 = 360. thus the last
equation is ∠𝐴𝐵𝐶 + ∠𝐵𝐶𝐷 = 180

9. Parallelograms
Diagonals bisect each other.
B
A D
C
Proof:
Given that □ABCD such that 𝐴𝑃 ≅ 𝐶𝑃 and 𝐵𝑃 ≅ 𝑃𝐷. We
to prove that □ABCD is a parallelogram.
By Vertical Angle Theorem , ∠𝐴𝑃𝐷 ≅ ∠𝐵𝑃𝐶. This implies
that ∆APD≅∆CPB by SAS≅ theo., and ∠𝐷𝐴𝑃 ≅ ∠𝐵𝐶𝑃 by
CPCTC. This means that 𝐴𝐷 ∥ 𝐵𝐶. Similarly, 𝐴𝐵 ∥ 𝐷𝐶.
Hence □ABCD is a parallelogram. QED
P

10. Parallelograms
Opposite sides are congruent.
Proof:
Given that □ABCD such that 𝐴𝐵 ≅ 𝐶𝐷 and 𝐵𝐶 ≅ 𝐴𝐷. We
to prove that □ABCD is a parallelogram.
Since 𝐴𝐵 ≅ 𝐶𝐷 and 𝐵𝐶 ≅ 𝐴𝐷, then we can say that
∆ABD≅∆CDB by SSS≅ theo. 𝐵𝐷 is transversal line 𝐴𝐷 and
𝐵𝐶, and ∠𝐷𝐵𝐶 ≅ ∠𝐴𝐷𝐵 (CPCTC) then 𝐴𝐵 ∥ 𝐵𝐶 by the
Converse of Alternate Interior Angles. Hence □ABCD is a
parallelogram. QED
B
A D
C

11. Parallelograms
The opposite sides are parallel
and congruent.
Proof:
Given that □ABCD such that 𝐴𝐷 ∥ 𝐵𝐶 and 𝐴𝐷 = 𝐵𝐶. We
to prove that □ABCD is a parallelogram.
The diagonal BD is a transversal of 𝐴𝐷 and 𝐵𝐶. And by
Alternate Interior angle theo., ∠𝐴𝐷𝐵 ≅ ∠𝐶𝐵𝐷. 𝐵𝐷 is a
common side between ∆ADB and ∆CBD which are
congruent by ASA. By CPCTC, AB=CD. Therefore, □ABCD
is a parallelogram.
B
A D
C

12. Parallelogram
Find the values of x and y in the parallelogram ABCD.
B
A
x0
C
y0
400
600
Answer: y = 1200 and x = 200

13. Parallelogram
In the figure below, □AECG is a parallelogram. If 𝐵𝐸 ≅
𝐺𝐷, prove that □ABCD is also a parallelogram.
A
B C
D
E
G

14. Statement Reason
1. Draw 𝐴𝐶 such that 𝐴𝐶 intersect 𝐸𝐺 at X 1. Construction
2. 𝐴𝐶 is diagonal to □AECG and □ABCD 2. Def. of diagonal line
3. 𝐸𝑋 ≅ 𝑋𝐺 3. Prop. of parallelogram
4. 𝐵𝐸 ≅ 𝐺𝐷 4. Given
5. 𝐵𝑋 ≅ 𝐵𝐸 + 𝐸𝑋; 𝐷𝑋 ≅ 𝑋𝐺 + 𝐺𝐷 5. Segment Addition Postulate
6. 𝐵𝑋 ≅ 𝐵𝐸 + 𝐸𝑋; 𝐷𝑋 ≅ 𝐸𝑋 + 𝐵𝐸 6. Substitution
7. 𝐵𝑋 ≅ 𝐷𝑋 7. Transitive PE
8. 𝐴𝑋 ≅ 𝑋𝐶 8. Prop of Parallelogram (□AECG)
9. ∴ □ABCD is a parallelogram 9. Prop of Parallelogram (7 and 8)
x

15. Kinds of Parallelogram
Rectangle – All angles are congruent
Rhombus – All sides are congruent
Square – All angles and sides are congruent

16. Theorems on Rectangle
1. In a rectangle, the two diagonals are congruent.
Statements Reasons
1. □ABCD is rectangle 1. Given
2. ∠𝐴𝐵𝐶 ≅ ∠𝐶𝐷𝐴 ≅ ∠𝐵𝐴𝐷 ≅ ∠𝐵𝐶𝐷 2. Def. of rectangle
3. 𝐴𝐵 ≅ 𝐷𝐶, 𝐵𝐶 ≅ 𝐴𝐷 3. Prop. Of rectangle
4. ∆ABC≅ ∆𝐶𝐷𝐴 ≅ ∆𝐴𝐵𝐷 ≅∆DCB 4. SAS ≅ Theorem
5. 𝐵𝐷 ≅ 𝐴𝐶 5. CPCTC
A D
C
B
Given: □ABCD is a rectangle
Prove: 𝐵𝐷 ≅ 𝐴𝐶

17. Theorems on Rectangle
2. If one angle of a parallelogram is right, then it is a
rectangle.
Proof:
Given that ∠𝐴 of parallelogram ABCD is right, we need to prove that the
parallelogram is rectangle.
Since the consecutive angles of a parallelogram are supplementary, we can say ∠𝐴
and ∠𝐵 900 since they are supplementary. From the properties of parallelogram
that the opposite angles are congruent, then ∠𝐴 and ∠𝐶 are congruent. And since
∠𝐵 is opposite to ∠𝐷. Therefore parallelogram ABCD is a rectangle.

18. Theorems on Rhombus
1. The diagonals of a rhombus are perpendicular and
bisect each other.
Given: 𝐴𝐶 and 𝐵𝐷 are diagonals intersect at P.
𝐴𝐶 ⊥ 𝐵𝐷, 𝐴𝑃 ≅ 𝑃𝐶, 𝐵𝑃 ≅ 𝑃𝐷.
Prove: □ABCD is a rhombus.
A
B C
D
P
Statements Reasons
1. 𝐴𝐶 ⊥ 𝐵𝐷 1. Given
2. ∠𝐴𝑃𝐵 ≅ ∠𝐶𝑃𝐷 ≅ ∠𝐶𝑃𝐵 ≅ ∠𝐴𝑃𝐷 2. Def. of ⊥
3. 𝐴𝑃 ≅ 𝑃𝐶, 𝐵𝑃 ≅ 𝑃𝐷 3. Given
4. ∆APB ≅ ∆𝐶𝑃𝐵 ≅ ∆𝐶𝑃𝐷 ≅∆APD 4. SAS ≅ Theorem
5. 𝐴𝐵 ≅ 𝐵𝐶 ≅ 𝐶𝐷 ≅ 𝐷𝐴 5. CPCTC
6. ∴ □ABCD is a rhombus 6. Prop of rhombus

19. Midline Theorem
The segment that joins the midpoints of two sides of a
triangle is parallel to the third side and half as long.
Given: In ∆ABC, P and Q are midpoints of
𝐴𝐵 and 𝐴𝐶, respectively.
Prove: 𝑃𝑄 ∥ 𝐵𝐶 and PQ=
1
2
𝐵𝐶.
A
P
B
Q
C
R
Proof: Let R be a point in 𝑃𝑄 such that 𝑃𝑄 ≅ 𝑄𝑅.
Since Q is midpoint 𝐴𝐶, 𝐴𝑃 ≅ 𝑃𝐵. And by Vertical Angle
Theom., ∠𝐴𝑄𝑃 ≅ ∠𝑅𝑄𝐶, so ∆AQP≅∆CQR by SAS cong. Theo.
which implies that ∠𝑃𝐴𝑄 ≅ ∠𝑄𝐶𝑅 and 𝐴𝑃 ≅ 𝑅𝐶 by CPCTC.
This means that □PBCR is parallelogram because
𝐴𝐵 ∥ 𝑅𝐶 (converse of AIA) and 𝑃𝐵 ≅ 𝑅𝐶 by transitivity PE.
Therefore 𝑃𝑄 ∥ 𝐵𝐶.

20. Midline Theorem
The segment that joins the midpoints of two sides of a
triangle is parallel to the third side and half as long.
Given: In ∆ABC, P and Q are midpoints of
𝐴𝐵 and 𝐴𝐶, respectively.
Prove: 𝑃𝑄 ∥ 𝐵𝐶 and PQ=
1
2
𝐵𝐶.
A
P
B
Q
C
R
Since □PBCR is parallelogram, 𝑃𝑅 ≅ 𝐵𝐶. We can say that
PR=PQ+QR by Segment Addition postulate. Since PQ=QR
and PR=BC, then BC=2PQ. Simplify, the results will be
PQ=
1
2
𝐵𝐶

21. Theorems on Trapezoids
1. The median of a trapezoid is half of the sum of its
bases.
2. The base angles of an isosceles trapezoid are
congruent.
3. Opposite angles of an isosceles trapezoid are
supplementary.
4. The diagonals of an isosceles trapezoid are
congruent.

22. Theorems on kite
1. The diagonals are perpendicular to each other.
2. The area is half of the product of its diagonals.