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Sides AB, AC and altitude AD of $\Delta ABC$ are proportional to sides PQ, PR and altitude PS of another triangle PQR. Prove that $\Delta ABC \sim \Delta PQR$.
OR
In the given figure, PQRS is a trapezium with $PQ \parallel SR$. 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 $\Delta PTS \sim \Delta QTR$.
OR
In the given figure, PQRS is a trapezium with $PQ \parallel SR$. 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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(a) For correct figure (1 Mark)
In $\Delta ABD$ and $\Delta PQS$,
$\frac{AB}{PQ} = \frac{AD}{PS}$ (given)
$\Rightarrow \frac{AD}{PS} = \frac{AB}{PQ}$
$\sin B = \sin Q$
$\angle B = \angle Q$ (1 Mark)
$\angle D = \angle S$ (each $90^\circ$) (1/2 Mark)
$\Delta ABD \sim \Delta PQS$ (by AA similarity)
$\angle BAD = \angle QPS$ -------(i) (1/2 Mark)
Similarly, $\angle DAC = \angle SPR$-------(ii)
Adding (i) and (ii) we get $\angle BAC = \angle QPR$
Also $\frac{AB}{PQ} = \frac{AC}{PR}$ (1 Mark)
$\Delta ABC \sim \Delta PQR$ (by SAS similarity)
OR
(b) In $\Delta PQT$ and $\Delta RST$,
$\angle TPQ = \angle TRS$ (alternate interior angle) (1/2 Mark)
$\angle PQT = \angle TSR$ (alternate interior angle) (1/2 Mark)
$\Delta PQT \sim \Delta RST$ (by AA similarity)
So $\frac{PT}{RT} = \frac{QT}{ST}$ (1/2 Mark)
$\Rightarrow PT \cdot ST = QT \cdot RT$
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)
In $\Delta ABD$ and $\Delta PQS$,
$\frac{AB}{PQ} = \frac{AD}{PS}$ (given)
$\Rightarrow \frac{AD}{PS} = \frac{AB}{PQ}$
$\sin B = \sin Q$
$\angle B = \angle Q$ (1 Mark)
$\angle D = \angle S$ (each $90^\circ$) (1/2 Mark)
$\Delta ABD \sim \Delta PQS$ (by AA similarity)
$\angle BAD = \angle QPS$ -------(i) (1/2 Mark)
Similarly, $\angle DAC = \angle SPR$-------(ii)
Adding (i) and (ii) we get $\angle BAC = \angle QPR$
Also $\frac{AB}{PQ} = \frac{AC}{PR}$ (1 Mark)
$\Delta ABC \sim \Delta PQR$ (by SAS similarity)
OR
(b) In $\Delta PQT$ and $\Delta RST$,
$\angle TPQ = \angle TRS$ (alternate interior angle) (1/2 Mark)
$\angle PQT = \angle TSR$ (alternate interior angle) (1/2 Mark)
$\Delta PQT \sim \Delta RST$ (by AA similarity)
So $\frac{PT}{RT} = \frac{QT}{ST}$ (1/2 Mark)
$\Rightarrow PT \cdot ST = QT \cdot RT$
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)