201
Sides AB, AC and altitude AD of $\Delta ABC$ are proportional to sides PQ, PR and altitude PS of another triangle PQR. Prove that $\Delta ABC \sim \Delta PQR$.
Show SolutionHide Solution↓
Solution (a)
In $\Delta ABD$ and $\Delta PQS$,
$\frac{AB}{PQ} = \frac{AD}{PS}$ (given) (1 Mark for correct figure)
$\Rightarrow \frac{AD}{AB} = \frac{PS}{PQ}$
$\sin B = \sin Q$
$\angle B = \angle Q$ (1 Mark)
$\angle D = \angle S$ (each $90^\circ$)
$\Delta ABD \sim \Delta PQS$ (by AA similarity) (1/2 Mark)
$\angle BAD = \angle QPS$ -------(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)
In $\Delta ABD$ and $\Delta PQS$,
$\frac{AB}{PQ} = \frac{AD}{PS}$ (given) (1 Mark for correct figure)
$\Rightarrow \frac{AD}{AB} = \frac{PS}{PQ}$
$\sin B = \sin Q$
$\angle B = \angle Q$ (1 Mark)
$\angle D = \angle S$ (each $90^\circ$)
$\Delta ABD \sim \Delta PQS$ (by AA similarity) (1/2 Mark)
$\angle BAD = \angle QPS$ -------(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)