2002 JBMO Problems/Problem 1

Problem

The triangle $ABC$ has $CA = CB$ . $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$ ). $D$ is the foot of the perpendicular from $C$ to $PB$ . Show that $PA + PB = 2 \cdot PD$ .

Solution

Since $APBC$ is a cyclic quadrilateral, $\angle CPA = \angle CBA$ , and $\angle ABP = \angle ACP$ --- (1)

Now let $E$ be the foot of the perpendicular from $C$ to $AP$ . Then we have, $CEPD$ is a cyclic quadrilateral with $CP$ as diameter of the circumcircle.

It follows that $\triangle CPE$ and $\triangle CDP$ are congruent (since $\angle CPA = \angle CPB$ ). So, we have $CE = CD$ and $PE = PD$ --- (2)

Also, in $\triangle CAE$ we have $\angle EAC = \angle CPA + \angle ACP = \angle CBA + \angle ABP$ ( from (1) above) Thus $\angle EAC = \angle DBC$ So from $SAS$ , $\triangle CAE$ is congruent to $\triangle CBD$ . Hence we have $AE = DB$ .

Now, $PA + PB = PA + PD + DB = PA + PD + AE = PE + PD = 2 . PD$ (from (2) above)

$Kris17$

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2002 JBMO Problems/Problem 1

Problem

Solution