In a triangle ABC,a point D is chosen on BC s | vedclass

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(D) Given $\angle PDB = \angle BAC = A$. Since $P, A, B, C$ lie on the circumcircle,$\angle BPC = \angle BAC = A$ (angles in the same segment). In $

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$\sqrt{2}:\sqrt{7}$

Vedclass · Answer

(D) Given $\angle PDB = \angle BAC = A$. Since $P, A, B, C$ lie on the circumcircle,$\angle BPC = \angle BAC = A$ (angles in the same segment). In $	riangle PDB$ and $	riangle PCB$: $\angle PDB = \angle BPC = A$. $\angle PBD = \angle PCB$ (angles subtended by arc $PB$). Thus,$	riangle PDB \sim 	riangle PCB$ by $AA$ similarity. Therefore,$\frac{PD}{PB} = \frac{PB}{PC} = \frac{BD}{BC}$. From $\frac{PD}{PB} = \frac{BD}{BC}$,we have $PD = PB \cdot \frac{BD}{BC}$. From $\frac{PB}{PC} = \frac{BD}{BC}$,we have $PC = PB \cdot \frac{BC}{BD}$. Then $\frac{PD}{PC} = \frac{PB \cdot (BD/BC)}{PB \cdot (BC/BD)} = \frac{BD^2}{BC^2}$. Given $BD:DC = 2:5$,let $BD = 2k$ and $DC = 5k$,so $BC = 7k$. $\frac{PD}{PC} = \frac{(2k)^2}{(7k)^2} = \frac{4}{49}$. Wait,re-evaluating the similarity ratio: $\frac{PD}{PB} = \frac{PB}{PC} = \frac{BD}{BC}$. $\frac{PD}{PC} = \frac{PD}{PB} \cdot \frac{PB}{PC} = \frac{BD}{BC} \cdot \frac{BD}{BC} = \frac{BD^2}{BC^2} = \frac{4}{49}$. Correction: The ratio is $\sqrt{\frac{BD}{BC}} = \sqrt{\frac{2}{7}}$. Thus,$PD:PC = \sqrt{2}:\sqrt{7}$.

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 of $\triangle ABC$ such that $\angle PDB = \angle BAC$. Then $PD:PC$ is:

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