Problem 4

In triangle

ABC

the bisector of angle

BCA

intersects the circumcircle again at

R

, the perpendicular bisector of

BC

P

, and the perpendicular bisector of

AC

Q

. The midpoint of

BC

K

and the midpoint of

AC

L

. Prove that the triangles

RPK

and

RQL

have the same area.

Step 2 of 4: Show triangle OPQ is isosceles at the circumcenter O

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In plain words

Because $O$ lies on both perpendicular bisectors $PK$ and $QL$ , the base angles of $\triangle OPQ$ along $CR$ match the equal acute angles from Step 1.

\angle OPQ = \angle CPK = \angle CQL = \angle OQP \implies OP = OQ

Detailed analysis

Let $O$ be the circumcenter of $\triangle ABC$ , which lies on both perpendicular bisectors $KP$ and $LQ$ . From $\angle CPK = \angle CQL$ (Step 1), the vertical/collinear angles at $P$ and $Q$ in $\triangle OPQ$ satisfy $\angle OPQ = \angle OQP$ , so $\triangle OPQ$ is isosceles with $OP = OQ$ .