Length relations

Geometry Level 3

An isosceles $\triangle ABC$ is such that $CA = CB$ . A point $P$ is on the circumcircle between $A$ and $B$ and on the opposite side of the line $AB$ to $C$ . Let $D$ be the foot of the perpendicular from $C$ to $PB$ . Given that $PB=3$ and $PA=5$ , find the length of $PD.$

The answer is 4.000.

1 solution

Sathvik Acharya
Jan 9, 2019

Construction: Extend $PB$ to point $Q$ such that $PA=BQ$ . Since $PACB$ is cyclic, $\angle PAC= \angle CBQ \implies \triangle CAP\cong \triangle CBQ$ as $CA=CB$ and $PA=BQ$ .

So, we have, $CP=CQ\implies \triangle PCQ$ isosceles. Also, $D$ is the midpoint of $PQ$ as $CD\perp PQ\implies PD=DQ$ .

Therefore, $\begin{aligned} PA+PB&=BQ+PB \\ &=PD+DQ \\ &=2PD \end{aligned}$ $\implies PD=\frac{PA+PB}{2}=\boxed{4}$

Length relations

The answer is 4.000.

1 solution

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