Let A, B, C, D be four points which determine the four-sided figure consisting of the four sides AB, BC, CD, and AD. Any four-sided figure in the plane is called a quadrilateral. If the opposite sides of the quadrilateral are parallel, that is, if AB is parallel to CD and AD is parallel to BC, and then the figure is called a parallelogram:

Figure 1. A parallelogram

Note:

Please write,

m(ABC) eq the size of angle ABC, and

d(P, Q) eq the length of segmen PQ.

**1. Parallelogram**

*Theorem.*

*1.1. In a parallelogram, the size of opposite angles is same.*

*Proof*.

Look at the figure below.

Figure 2. A parallelogram with its some parts

It is given a parallelogram ABCD.

Prove that m(A) = m(C) and m(B) = m(D).

Draw diagonal AC.

Obvious

m(A1) = m(C1) (because of alternate interior angles) and

m(A2) = m(C2) (because of alternate interior angles).

We get m(A1) + m(A2) = m(C1) + m(C2)

eq m(A) = m(C).

By the similar step, we get m(B) = m(D).

So the size of opposite angles in a parallelogram is same.

*Theorem.*

*1.2. In a parallelogram, the length of opposite sides is same.*

Proof.

Look at the figure 2.

It is given a parallelogram ABCD.

Prove that AB = CD and BC = DA.

Draw diagonal AC.

Look at DABC and DACD

Obvious

m(A1) = m(C1) (because of alternate interior angles),

m(A2) = m(C2) (because of alternate interior angles), and

AC = AC (because of coincide).

We get triangle ABC kongruent with triangle ACD (because of ASA)

One of the consequence is AB = CD and BC = AD.

*Theorem.*

*1.3. In a parallelogram, both of its diagonals are intersecting in the midpoint of its diagonals.*

Proof.

Look at figure below.

Figure 3. A parallelogram with its some parts

It is given a parallelogram ABCD.

Prove that AT = TC and BT = TD.

Look at triangle ATD and triangle BTC.

Obvious

m(D2) = m(B2) (because of alternate interior angles),

m(A2) = m(C2) (because of alternate interior angles), and

AD = BC (

**).***why?*We get triangle ATD congruent with triangle BTC (because of ASA)

One of the consequence is AT = TC and BT = TD.

So that T is midpoint of AC and T is also midpoint of BD.

*Theorem.*

*1.4. In a quadrilateral, if the size of opposite angles is same, then the quadrilateral given is a parallelogram.*

Proof.

Look at the figure 1.

It is given a quadrilateral ABCD and

m(A) = m(C) also m(B) = m(D).

Prove that ABCD is a parallelogram.

Obvious the sum of all angles in 4-gon is (4 – 2).180

^{o}= 360^{o}Obvious

m(A) + m(B) + mÐC + mÐD = 360

m(A) + m(B) + mÐC + mÐD = 360

^{o}eq m(A) + m(B) + m(A) + m(B) = 360

^{o}eq 2. m(A) + 2.m(B) = 360

^{o}eq m(A) + m(B) = 180

^{o}.We get:

1. Because of m(A) + m(B) = 180

^{o}then AD // BC.2. Because of m(A) + m(B) = 180

^{o}and m(B) = m(D) then m(A) + m(D) = 180^{o}.3. Because of m(A) + m(D) = 180

^{o}then AB // CD.Because AD // BC and AB // CD, we conclude that quadrilateral ABCD is a parallelogram.

*Theorems.*

*1.5. In a quadrilateral, if the length of opposite sides is same, then the quadrilateral given is a parallelogram.*

*1.6. In a quadrilateral, if both of diagonals are intersecting one another at the midpoint, then the quadrilateral given is a parallelogram.*

*1.7. In a quadrilateral, if there are two pairs parallel sides, then quadrilateral given is a parallelogram.*

Proof.

We leave the proof for 1.5, 1.6, and 1.7 as exercises.

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