Question

use the given information to prove that $\triangle sqr \cong \triangle qsp$.
given: $\triangle sqr$ and $\triangle qsp$ are right triangles
$\overline{pq} \cong \overline{rs}$
prove: $\triangle sqr \cong \triangle qsp$
step 1: $\triangle sqr$ and $\triangle qsp$ are right triangles (given)
step 2: $\overline{pq} \cong \overline{rs}$ (given)
step 3: $\triangle sqr \cong \triangle qsp$ (reason?)

Explanation:

Step 1: Identify the right triangles

We know that \(\triangle SQR\) and \(\triangle QSP\) are right triangles (given). So, \(\angle R=\angle P = 90^{\circ}\).

Step 2: Identify the hypotenuse and a leg

We are given that \(\overline{PQ}\cong\overline{RS}\). Also, the hypotenuse \(\overline{SQ}\) is common to both \(\triangle SQR\) and \(\triangle QSP\), so \(\overline{SQ}\cong\overline{SQ}\) (by the reflexive property of congruence).

Step 3: Apply the Hypotenuse - Leg (HL) theorem

For right triangles, if the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. Here, hypotenuse \(\overline{SQ}\cong\overline{SQ}\) and leg \(\overline{PQ}\cong\overline{RS}\), so by HL theorem, \(\triangle SQR\cong\triangle QSP\).

Answer:

The reason for step 3 is "Hypotenuse - Leg (HL) Congruence Theorem (since \(\triangle SQR\) and \(\triangle QSP\) are right triangles, \(\overline{SQ}\cong\overline{SQ}\) (reflexive), and \(\overline{PQ}\cong\overline{RS}\) (given), so by HL, \(\triangle SQR\cong\triangle QSP\))"