Question

quick proof sketch: if ∠p and ∠q are right angles, and ∠p and ∠r are supplementary, prove ∠q ≅ ∠r. (just outline steps, not full proof.)

Explanation:

Step1: State angle - measure of right angle

Since $\angle P$ and $\angle Q$ are right - angles, $m\angle P=m\angle Q = 90^{\circ}$.

Step2: Use supplementary - angle property

Since $\angle P$ and $\angle R$ are supplementary, $m\angle P+m\angle R=180^{\circ}$. Substituting $m\angle P = 90^{\circ}$ into the equation, we get $90^{\circ}+m\angle R=180^{\circ}$, so $m\angle R = 90^{\circ}$.

Step3: Compare angle measures

Since $m\angle Q = 90^{\circ}$ and $m\angle R = 90^{\circ}$, then $\angle Q\cong\angle R$ (angles with equal measures are congruent).

Answer:

$\angle Q\cong\angle R$ is proved by showing $m\angle Q=m\angle R = 90^{\circ}$ through the properties of right - angles and supplementary angles.