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S is any point on the side QR of a $\triangle PQR$ such that $\angle PSR = \angle QPR$. Prove that $\frac{QR}{RP} = \frac{RP}{RS}$.
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Solution:
In $\triangle PQR$ and $\triangle SPR$
$\angle QPR = \angle PSR$ (Given) (1 Mark)
$\angle R = \angle R$ (common) (1 Mark)
$\Rightarrow \triangle PQR \sim \triangle SPR$ (AA similarity)
$\therefore \frac{QR}{PR} = \frac{PR}{SR}$ (1 Mark)
or $\frac{QR}{RP} = \frac{RP}{RS}$
In $\triangle PQR$ and $\triangle SPR$
$\angle QPR = \angle PSR$ (Given) (1 Mark)
$\angle R = \angle R$ (common) (1 Mark)
$\Rightarrow \triangle PQR \sim \triangle SPR$ (AA similarity)
$\therefore \frac{QR}{PR} = \frac{PR}{SR}$ (1 Mark)
or $\frac{QR}{RP} = \frac{RP}{RS}$