Question

In $△PQR,\angle PQR={90}^o,QD\perp PR$.
If $PD=4DR$. Prove that $PQ=2QR$.

Answer 1

It is given that $\angle PQR={90}^0$, $QS\perp PR$, $PD=4DR$

Consider,

$P{Q}^2=PR\times PDPR=\frac{P{Q}^2}{PD}.......\left(1\right)$

Similarly, we have,

$Q{R}^2=PR\times RDPR=\frac{Q{R}^2}{RD}.......\left(2\right)$

Comparing equations 1 and 2 as follows:

$\frac{P{Q}^2}{PD}=\frac{Q{R}^2}{RD}\Rightarrow \frac{P{Q}^2}{4DR}=\frac{Q{R}^2}{RD}\Rightarrow \frac{P{Q}^2}{4}=Q{R}^2\Rightarrow P{Q}^2=4Q{R}^2\Rightarrow \sqrt{P{Q}^2}=\sqrt{4Q{R}^2}\Rightarrow PQ=2QR$

Hence, $PQ=2QR$.