Vertically opposite angles

When two lines intersect they form four angles, in pairs opposite with respect to the point of meeting. The first application of the congruence criteria is to prove that vertically opposite angles are equal.

Theorem — Vertically opposite angles

Two vertically opposite angles are congruent.

Proof

Let $\widehat{CMA}$ and $\widehat{BMD}$ be vertically opposite angles formed by the intersection of two lines at $M$. The idea is to show that the two angles appear in congruent triangles.

1. Hypothesis. $\widehat{CMA}$ and $\widehat{BMD}$ are vertically opposite (built by extension).
2. I consider. The triangles $CMA$ and $BMD$ and I compare their elements: $\begin{array}{|c|c|} \hline \widehat{CMA} & \widehat{BMD} \\ \hline CM & BM \\ AM & DM \\ \hline \end{array}$
3. I deduce. It follows that $\triangle CMA \cong \triangle BMD$ by the first criterion.
4. Verified. The corresponding elements of the two triangles are therefore congruent: $AC\cong DB,\quad \widehat{CAM}\cong\widehat{BDM},\quad \boxed{\widehat{ACM}\cong\widehat{DBM}}$

$\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Vertically opposite angles · Angle · Congruence criteria · Proof
Skills: Proving · Synthetic geometry