In $ riangle ABC$ and $ riangle PQR$, $AB = PQ$, $BC = QR$, and $\angle B = \angle Q = 90^\circ$. Which congruence criterion proves that $ riangle ABC \cong riangle PQR$?

Question

In $	riangle ABC$ and $	riangle PQR$, $AB = PQ$, $BC = QR$, and $\angle B = \angle Q = 90^\circ$. Which congruence criterion proves that $	riangle ABC \cong 	riangle PQR$?

Accepted Answer

Given $AB = PQ$ and $BC = QR$, the included angle between sides $AB$ and $BC$ is $\angle B$, and between sides $PQ$ and $QR$ is $\angle Q$. Since $\angle B = \angle Q = 90^\circ$, two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle. Therefore, by the SAS (Side-Angle-Side) congruence criterion, $\triangle ABC \cong \triangle PQR$.

Note: RHS criterion requires the hypotenuse and one side to be equal, but hypotenuse AC = PR is not given here.

MP Board · Class 9 · Mathematics · TrianglesIn $\triangle ABC$ and $\triangle PQR$, $AB = PQ$, $BC = QR$, and $\angle B = \angle Q = 90^\circ$. Which congruence criterion proves that $\triangle ABC \cong \triangle PQR$?