proof of parallelogram theorems

proof of parallelogram theorems

Theorem 1

.

The opposite sides of a

parallelogram

are

congruent

.

Proof.

This was proved in the parent (

http://planetmath.org/ParallelogramTheorems

) article.
∎

Theorem 2

.

If both pairs of opposite sides of a

quadrilateral

are congruent, the quadrilateral is a parallelogram.

Proof.

Suppose

A

⁢

B

⁢

C

⁢

D

is the given parallelogram, and draw

A

⁢

C

¯

.

Then

△

⁢

A

⁢

B

⁢

C

≅

△

⁢

A

⁢

D

⁢

C

by SSS, since by

assumption

A

⁢

B

=

C

⁢

D

and

A

⁢

D

=

B

⁢

C

, and the two

triangles

share a third side.

By CPCTC, it follows that

∠

⁢

B

⁢

A

⁢

C

≅

∠

⁢

D

⁢

C

⁢

A

and that

∠

⁢

B

⁢

C

⁢

A

≅

∠

⁢

D

⁢

A

⁢

C

. But the

theorems

about corresponding angles in

transversal

cutting then imply that

A

⁢

B

¯

and

C

⁢

D

¯

are

parallel

, and that

A

⁢

D

¯

and

B

⁢

C

¯

are parallel. Thus

A

⁢

B

⁢

C

⁢

D

is a parallelogram.
∎

Theorem 3

.

If one pair of opposite sides of a quadrilateral are both parallel and congruent, the quadrilateral is a parallelogram.

Proof.

Again let

A

⁢

B

⁢

C

⁢

D

be the given parallelogram. Assume

A

⁢

B

=

C

⁢

D

and that

A

⁢

B

¯

and

C

⁢

D

¯

are parallel, and draw

A

⁢

C

¯

.

Since

A

⁢

B

¯

and

C

⁢

D

¯

are parallel, it follows that the alternate interior angles are equal:

∠

⁢

B

⁢

A

⁢

C

≅

∠

⁢

D

⁢

C

⁢

A

. Then by SAS,

△

⁢

A

⁢

B

⁢

C

≅

△

⁢

A

⁢

D

⁢

C

since they share a side.

Again by CPCTC we have that

B

⁢

C

=

A

⁢

D

, so both pairs of sides of the quadrilateral are congruent, so by Theorem

2

, the quadrilateral is a parallelogram.
∎

Theorem 4

.

The

diagonals

of a parallelogram bisect each other.

Proof.

Let

A

⁢

B

⁢

C

⁢

D

be the given parallelogram, and draw the diagonals

A

⁢

C

¯

and

B

⁢

D

¯

, intersecting at

E

.

Since

A

⁢

B

⁢

C

⁢

D

is a parallelogram, we have that

A

⁢

B

=

C

⁢

D

. In

addition

,

A

⁢

B

¯

and

C

⁢

D

¯

are parallel, so the alternate interior angles are equal:

∠

⁢

A

⁢

B

⁢

D

≅

∠

⁢

B

⁢

D

⁢

C

and

∠

⁢

B

⁢

A

⁢

C

≅

∠

⁢

A

⁢

C

⁢

D

. Then by ASA,

△

⁢

A

⁢

B

⁢

E

≅

△

⁢

C

⁢

D

⁢

E

.

By CPCTC we see that

A

⁢

E

=

C

⁢

E

and

B

⁢

E

=

D

⁢

E

, proving the theorem.
∎

Theorem 5

.

If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

Proof.

Let

A

⁢

B

⁢

C

⁢

D

be the given quadrilateral, and let its diagonals

intersect

in

E

.

Then by assumption,

A

⁢

E

=

E

⁢

C

and

D

⁢

E

=

E

⁢

B

. But also

vertical angles

are equal, so

∠

⁢

A

⁢

E

⁢

D

≅

∠

⁢

A

⁢

E

⁢

B

and

∠

⁢

C

⁢

E

⁢

D

≅

∠

⁢

A

⁢

E

⁢

B

. Thus, by SAS we have that

△

⁢

A

⁢

E

⁢

D

≅

△

⁢

C

⁢

E

⁢

B

and

△

⁢

C

⁢

E

⁢

D

≅

△

⁢

A

⁢

E

⁢

B

.

By CPCTC it follows that

A

⁢

B

=

C

⁢

D

and that

A

⁢

D

=

B

⁢

C

. By Theorem

1

,

A

⁢

B

⁢

C

⁢

D

is a parallelogram.

∎

Title

proof of parallelogram theorems

Canonical name

ProofOfParallelogramTheorems

Date of creation

2013-03-22 17:15:48

Last modified on

2013-03-22 17:15:48

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

9

Author

rm50 (10146)

Entry type

Proof

Classification

msc 51M04

Classification

msc 51-01

Related topic

Parallelogram

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