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  4. 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 A B C such that angle P D B= angle B A C. Then PD: PC is :-

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Q. 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 $A B C$ such that $\angle P D B=\angle B A C$. Then $PD : PC$ is :-

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Solution:

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$\frac{ BD }{ DC }=\frac{2}{5}$
$\angle PDB =\angle BAC =\theta$
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 $\triangle PCB \sim \triangle PDB$
$\frac{P C}{D P}=\frac{B C}{P B}=\frac{P B}{B D}$
$\left(\frac{P C}{D P}\right)^{2}=\frac{B C}{P B} \times \frac{P B}{B D}=\frac{B C}{B D}$
$\frac{P C}{D P}=\sqrt{\frac{B C}{B D}}=\sqrt{\frac{7 \lambda}{2 \lambda}}=\frac{\sqrt{7}}{\sqrt{2}} $
$\frac{D P}{P C}=\frac{\sqrt{2}}{\sqrt{7}}$