Parallelogram Opposite Sides Theorem

This theorem can be proven by placing the parallelogram on a coordinate plane. For simplicity, vertex P will be placed at the origin and vertex S on the x-axis.

Since P lies on the origin, its coordinates are (0,0). Point S is on the x-axis, meaning its y-coordinate is 0. Let a be the x-coordinate of S. Furthermore, let b and c be the coordinates of Q. P(0,0) Q(b,c) S(a,0) Note that both P and S lie on the x-axis. Therefore, SP is a horizontal segment. Since opposite sides of a parallelogram are parallel, QR is also a horizontal segment. This means that Q and R have the same y-coordinate. Let x be the x-coordinate of R.

Next, the x-coordinate of R will be determined. Since PQ and SR are parallel, they have the same slope. The slope of PQ can be found using the Slope Formula. The slope of PQ is cb. By following the same procedure, the slope of SR can be expressed in terms of x.

m = y_2-y_1/x_2-x_1
Side	Endpoints	Substitute	Simplify
PQ	P( 0, 0) and Q( b, c)	m_(PQ)=c- 0/b- 0	m_(PQ)=c/b
SR	S( a, 0) and R( x, c)	m_(SR)=c- 0/x- a	m_(SR)=c/x-a

As it has been previously stated, since PQ and SR are parallel, their slopes are equal. m_(PQ)=m_(SR) ⇕ c/b=c/x-a The above equation can be solved for x. The x-coordinate of R is a+b. Finally, by using the Distance Formula, the length of each side of the parallelogram can be calculated. The length of PQ will be calculated first. By following the same procedure, all the side lengths can be calculated.

d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
Side	Endpoints	Substitute	Simplify
PQ	P( 0, 0) and Q( b, c)	PQ= sqrt(( b- 0)^2+( c- 0)^2)	PQ=sqrt(b^2+c^2)
QR	Q( b, c) and R( a+b, c)	QR= sqrt(( a+b- b)^2+( c- c)^2)	QR=a
SR	S( a, 0) and R( a+b, c)	SR= sqrt(( a+b- a)^2+( c- 0)^2)	SR=sqrt(b^2+c^2)
PS	P( 0, 0) and S( a, 0)	PS= sqrt(( a- 0)^2+( 0- 0)^2)	PS=a

By the Transitive Property of Equality, it can be said that PQ=SR and that QR=PS. cc PQ=sqrt(b^2+c^2) SR=sqrt(b^2+c^2) & QR=a PS=a ⇓ & ⇓ PQ=SR & QR=PS By definition of congruent segments, it can be stated that the opposite sides of a parallelogram are congruent.

This theorem can also be proven by using congruent triangles. Consider the parallelogram PQRS and its diagonal PR.

It can be noted that two triangles are formed with PR as a common side. △ PQR and △ RSP By the definition of a parallelogram, PQ and SR are parallel. Therefore, by the Alternate Interior Angles Theorem, it can be stated that ∠ QPR ≅ ∠ SRP and that ∠ QRP ≅ ∠ SPR. Furthermore, by the Reflexive Property of Congruence, PR is congruent to itself.

Consequently, △ PQR and △ RSP have two pairs of congruent angles and an included congruent side. ∠ QPR &≅ ∠ SRP PR&≅PR ∠ QRP &≅ ∠ SPR Therefore, by the Angle-Side-Angle Congruence Theorem, △ PQR and △ RSP are congruent triangles. △ PQR ≅ △ RSP Since corresponding parts of congruent figures are congruent, PS is congruent to QR and PQ is congruent to RS.