Question

In a triangle $ABC$, a point $D$ is chosen on $BC$ such that $BD:DC=2:5$. Let $P$ be a point on the circumcircle $ABC$ such that $\angle PDB=\angle BAC$. Then $PD:PC$ is :-

• A. $\sqrt{2}:\sqrt{5}$
• B. $2:5$
• C. $2:7$
• D. $\sqrt{2}:\sqrt{7}$

Answer 1

Answer: (D)

$\frac{BD}{DC}=\frac{2}{5}$
$\angle PDB=\angle BAC=\theta$
let $\angle PCD=\alpha$
$\Rightarrow \angle DPC=\theta -\alpha$
$\angle BAC=\angle BPC=\theta$
(angle in the same segment)
$\Rightarrow BPD=\theta -(\theta -\alpha )=\alpha$
so $△PCB\sim △PDB$
$\frac{PC}{DP}=\frac{BC}{PB}=\frac{PB}{BD}$
${\left(\frac{PC}{DP}\right)}^2=\frac{BC}{PB}\times \frac{PB}{BD}=\frac{BC}{BD}$
$\frac{PC}{DP}=\sqrt{\frac{BC}{BD}}=\sqrt{\frac{7\lambda }{2\lambda }}=\frac{\sqrt{7}}{\sqrt{2}}$
$\frac{DP}{PC}=\frac{\sqrt{2}}{\sqrt{7}}$