Parallelogram and some of its theorems

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

Definition.
Parallelogram is a quadrilateral which its corresponding sides are parallel.


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).180o = 360o
Obvious
m(A) + m(B) + mÐC + mÐD = 360o
eq m(A) + m(B) + m(A) + m(B) = 360o
eq 2. m(A) + 2.m(B) = 360o
eq m(A) + m(B) = 180o.
We get:
1.    Because of m(A) + m(B) = 180o then AD // BC.
2.    Because of m(A) + m(B) = 180o and m(B) = m(D) then m(A) + m(D) = 180o.
3.    Because of m(A) + m(D) = 180o 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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