Congruent opposite angles and parallelogram

We prove one of the sufficient conditions for the parallelogram: if the opposite angles are congruent, the quadrilateral is a parallelogram. The proof works on the sum of the interior angles split along a diagonal.

Theorem — Congruent opposite angles

If a quadrilateral has congruent opposite angles, then it is a parallelogram.

Proof

The proof proceeds by splitting the angles along the diagonal and exploiting the relations between the sums.

1. Hypothesis. $\widehat{A}\cong\widehat{C}$ and $\widehat{B}\cong\widehat{D}$.
2. Construction. We draw the diagonal $AC$, which divides $\widehat{A}$ into $\alpha,\gamma$ and $\widehat{C}$ into $\beta,\delta$.
3. Consider. From the sum of the interior angles: $(\alpha+\gamma)+\widehat{B}+(\beta+\delta)+\widehat{D}=2\pi$. Since $\widehat{B}\cong\widehat{D}$: $\alpha+\gamma+\beta+\delta+2\widehat{B}=2\pi$. In the triangle $ABC$: $\alpha+\widehat{B}+\beta=\pi$, hence $\gamma+\delta+\widehat{B}=\pi$. In the triangle $ACD$: $\gamma+\widehat{D}+\delta=\pi$, confirmed.
4. Key idea. From $\widehat{A}\cong\widehat{C}$: $\alpha+\gamma=\beta+\delta$. Comparing with $\alpha+\beta=\gamma+\delta$ (the difference of the triangle equations): we obtain $\alpha=\delta$ and $\beta=\gamma$.
5. Deduce. $\beta\cong\gamma \implies AB\parallel CD$ (alternate interior angles with transversal $AC$). $\alpha\cong\delta \implies BC\parallel AD$.
6. Verified. Hence $ABCD$ is a parallelogram.

$\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Alternate interior angles · Proof · Parallelogram · Sum of polygon angles
Skills: Proving · Synthetic geometry