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In the given figure, PQRS is a trapezium with PQ $||$ SR. Diagonals PR and QS intersect each other at a point T. Prove that $\frac{PT}{TR} = \frac{QT}{TS}$. Further, if TS = TR, prove that $\Delta PTS \sim \Delta QTR$.
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(b) In $\Delta PQT$ and $\Delta RST$,
$\angle TPQ = \angle TRS$ (alternate interior angles)
$\angle PQT = \angle TSR$ (alternate interior angles) (1/2 Mark)
$\Delta PQT \sim \Delta RST$ (by AA similarity) (1/2 Mark)
So $\frac{PT}{RT} = \frac{QT}{ST}$ (1/2 Mark)
$\Rightarrow PT = QT$ (RT = ST) (1/2 Mark)
which gives $\frac{PT}{QT} = 1$ and $\frac{ST}{RT} = 1$ (1 Mark)
Now In $\Delta PTS$ and $\Delta QTR$
$\frac{PT}{QT} = \frac{ST}{RT}$ (1/2 Mark)
$\angle PTS = \angle QTR$ (vertically opposite angles) (1/2 Mark)
$\Delta PTS \sim \Delta QTR$ (by SAS similarity) (1/2 Mark)
(b) In $\Delta PQT$ and $\Delta RST$,
$\angle TPQ = \angle TRS$ (alternate interior angles)
$\angle PQT = \angle TSR$ (alternate interior angles) (1/2 Mark)
$\Delta PQT \sim \Delta RST$ (by AA similarity) (1/2 Mark)
So $\frac{PT}{RT} = \frac{QT}{ST}$ (1/2 Mark)
$\Rightarrow PT = QT$ (RT = ST) (1/2 Mark)
which gives $\frac{PT}{QT} = 1$ and $\frac{ST}{RT} = 1$ (1 Mark)
Now In $\Delta PTS$ and $\Delta QTR$
$\frac{PT}{QT} = \frac{ST}{RT}$ (1/2 Mark)
$\angle PTS = \angle QTR$ (vertically opposite angles) (1/2 Mark)
$\Delta PTS \sim \Delta QTR$ (by SAS similarity) (1/2 Mark)