105
In the given figure, two medians PD and QE of $\triangle PQR$ meet each other at O. Prove that :
(i) $\triangle POQ \sim \triangle DOE$
(ii) $PO = 2OD$
(iii) $PO = \frac{2}{3} PD$
(i) $\triangle POQ \sim \triangle DOE$
(ii) $PO = 2OD$
(iii) $PO = \frac{2}{3} PD$
Show SolutionHide Solution↓
(i) As D and E are the mid-points of RQ and RP respectively.
By mid-point theorem, $ED \parallel PQ$ and $ED = \frac{1}{2} PQ$ ... (1)
$\Rightarrow \triangle POQ \sim \triangle DOE$
(ii) Using part (i), $\frac{PO}{OD} = \frac{PQ}{ED}$
Using (1), $PO = 2 OD$
(iii) Using part (ii), $PO = 2 OD = 2(PD – PO)$
$\Rightarrow 3PO = 2PD$
$\Rightarrow PO = \frac{2}{3} PD$
By mid-point theorem, $ED \parallel PQ$ and $ED = \frac{1}{2} PQ$ ... (1)
$\Rightarrow \triangle POQ \sim \triangle DOE$
(ii) Using part (i), $\frac{PO}{OD} = \frac{PQ}{ED}$
Using (1), $PO = 2 OD$
(iii) Using part (ii), $PO = 2 OD = 2(PD – PO)$
$\Rightarrow 3PO = 2PD$
$\Rightarrow PO = \frac{2}{3} PD$