Question

In ΔPQR, PQ = PR. QR is extended to a point S. Prove that PS > PQ.

Answer 1

Given:

• In ΔPQR, PQ = PR.
• QR is extended to a point S.

To Prove: PS > PQ

In ΔPQR:

Since PQ = PR, the angles opposite these equal sides are also equal.

Therefore, ∠PRQ = ∠PQR.

Consider ΔPQS:

In ΔPQS, ∠PQR is the exterior angle to the triangle PSR.

By the exterior angle theorem, the exterior angle of a triangle is greater than either of the opposite interior angles.

Therefore, ∠PQR (or PQS) > PSR.

Combine statements 3 and 4: 

Since ∠PQR > ∠PSR and ∠PRQ = ∠PQR, we can conclude that ∠PRQ > ∠PSR.

In ΔPSR:

The side opposite to ∠PRQ is PS and the side opposite to ∠PSR is PR.

In any triangle, the side opposite the greater angle is greater.

Since ∠PRQ > ∠PSR, we have PS > PR.

Final step: 

Given that PQ = PR, we can substitute PQ for PR in the inequality.

Therefore, PS > PQ.