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 $\\overline{AC}$ .

Then $\\triangle ABC\\cong\\triangle ADC$ by SSS, since by assumption $AB=CD$ and $AD=BC$ , and the two triangles share a third side.

By CPCTC, it follows that $\\angle BAC\\cong\\angle DCA$ and that $\\angle BCA\\cong\\angle DAC$ . But the theorems about corresponding angles in transversal cutting then imply that $\\overline{AB}$ and $\\overline{CD}$ are parallel, and that $\\overline{AD}$ and $\\overline{BC}$ 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 $AB=CD$ and that $\\overline{AB}$ and $\\overline{CD}$ are parallel, and draw $\\overline{AC}$ .

Since $\\overline{AB}$ and $\\overline{CD}$ are parallel, it follows that the alternate interior angles are equal: $\\angle BAC\\cong\\angle DCA$ . Then by SAS, $\\triangle ABC\\cong\\triangle ADC$ since they share a side.

Again by CPCTC we have that $BC=AD$ , 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 $\\overline{AC}$ and $\\overline{BD}$ , intersecting at $E$ .

Since $A B C D$ is a parallelogram, we have that $AB=CD$ . In addition, $\\overline{AB}$ and $\\overline{CD}$ are parallel, so the alternate interior angles are equal: $\\angle ABD\\cong\\angle BDC$ and $\\angle BAC\\cong\\angle ACD$ . Then by ASA, $\\triangle ABE\\cong\\triangle CDE$ .

By CPCTC we see that $AE=CE$ and $BE=DE$ , 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, $AE=EC$ and $DE=EB$ . But also vertical angles are equal, so $\\angle AED\\cong\\angle AEB$ and $\\angle CED\\cong\\angle AEB$ . Thus, by SAS we have that $\\triangle AED\\cong\\triangle CEB$ and $\\triangle CED\\cong\\triangle AEB$ .

By CPCTC it follows that $AB=CD$ and that $AD=BC$ . By Theorem 1, $A B C D$ is a parallelogram.

∎

单词	ProofOfParallelogramTheorems
释义	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 $\\overline{AC}$ . Then $\\triangle ABC\\cong\\triangle ADC$ by SSS, since by assumption $AB=CD$ and $AD=BC$ , and the two triangles share a third side. By CPCTC, it follows that $\\angle BAC\\cong\\angle DCA$ and that $\\angle BCA\\cong\\angle DAC$ . But the theorems about corresponding angles in transversal cutting then imply that $\\overline{AB}$ and $\\overline{CD}$ are parallel, and that $\\overline{AD}$ and $\\overline{BC}$ 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 $AB=CD$ and that $\\overline{AB}$ and $\\overline{CD}$ are parallel, and draw $\\overline{AC}$ . Since $\\overline{AB}$ and $\\overline{CD}$ are parallel, it follows that the alternate interior angles are equal: $\\angle BAC\\cong\\angle DCA$ . Then by SAS, $\\triangle ABC\\cong\\triangle ADC$ since they share a side. Again by CPCTC we have that $BC=AD$ , 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 $\\overline{AC}$ and $\\overline{BD}$ , intersecting at $E$ . Since $A B C D$ is a parallelogram, we have that $AB=CD$ . In addition, $\\overline{AB}$ and $\\overline{CD}$ are parallel, so the alternate interior angles are equal: $\\angle ABD\\cong\\angle BDC$ and $\\angle BAC\\cong\\angle ACD$ . Then by ASA, $\\triangle ABE\\cong\\triangle CDE$ . By CPCTC we see that $AE=CE$ and $BE=DE$ , 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, $AE=EC$ and $DE=EB$ . But also vertical angles are equal, so $\\angle AED\\cong\\angle AEB$ and $\\angle CED\\cong\\angle AEB$ . Thus, by SAS we have that $\\triangle AED\\cong\\triangle CEB$ and $\\triangle CED\\cong\\triangle AEB$ . By CPCTC it follows that $AB=CD$ and that $AD=BC$ . By Theorem 1, $A B C D$ is a parallelogram. ∎
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