We are given two isosceles triangles: $\triangle PQR$ with $PQ = PR$, and $\triangle QSR$ with $QS = SR$. They share a common base $QR$. We need to show that $PS$ bisects $QR$ at a right angle.
Let’s denote the intersection point of $PS$ and $QR$ as $M$. We want to prove that $QM = MR$ and $\angle PMS = \angle PMR = 90^\circ$.
Proof:
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Consider triangles $\triangle PQS$ and $\triangle PRS$:
- $PQ = PR$ (given)
- $QS = SR$ (given)
- $PS = PS$ (common side)
- Therefore, by SSS congruence, $\triangle PQS \cong \triangle PRS$.
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Corresponding parts of congruent triangles are congruent (CPCTC):
- Thus, $\angle QPS = \angle RPS$.
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Consider triangles $\triangle PQM$ and $\triangle PRM$:
- $PQ = PR$ (given)
- $\angle QPM = \angle RPM$ (from step 2, $\angle QPS=\angle RPS$)
- $PM = PM$ (common side)
- Therefore, by SAS congruence, $\triangle PQM \cong \triangle PRM$.
-
CPCTC again:
- $QM = MR$ (so, $PS$ bisects $QR$)
- $\angle PMQ = \angle PMR$
-
Since $\angle PMQ$ and $\angle PMR$ are a linear pair:
- $\angle PMQ + \angle PMR = 180^\circ$
- Since $\angle PMQ = \angle PMR$, we have $2\angle PMQ = 180^\circ$
- Thus, $\angle PMQ = \angle PMR = 90^\circ$. (so, $PS$ is perpendicular to $QR$)
