205
AD and PS are respectively, the medians of $\triangle ABC$ and $\triangle PQR$. If $\triangle ABC \sim \triangle PQR$, then prove that
(i) $\triangle ADC \sim \triangle PSR$
(ii) $\frac{AD}{PS} = \frac{BC}{QR}$
(i) $\triangle ADC \sim \triangle PSR$
(ii) $\frac{AD}{PS} = \frac{BC}{QR}$
Show SolutionHide Solution↓
OR
(b) (i) $\triangle ABC \sim \triangle PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} = \frac{2DC}{2SR}$ (2)
$\Rightarrow \frac{AC}{PR} = \frac{DC}{SR}$ (
frac{1}{2})
and $\angle C = \angle R$ (
frac{1}{2})
$\therefore \triangle ADC \sim \triangle PSR$ (by SAS similarity) (1)
(ii) Since $\triangle ADC \sim \triangle PSR$, $\frac{AD}{PS} = \frac{DC}{SR}$ (1)
$\frac{AD}{PS} = \frac{BC/2}{QR/2} = \frac{BC}{QR}$
(b) (i) $\triangle ABC \sim \triangle PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} = \frac{2DC}{2SR}$ (2)
$\Rightarrow \frac{AC}{PR} = \frac{DC}{SR}$ (
frac{1}{2})
and $\angle C = \angle R$ (
frac{1}{2})
$\therefore \triangle ADC \sim \triangle PSR$ (by SAS similarity) (1)
(ii) Since $\triangle ADC \sim \triangle PSR$, $\frac{AD}{PS} = \frac{DC}{SR}$ (1)
$\frac{AD}{PS} = \frac{BC/2}{QR/2} = \frac{BC}{QR}$