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 1 of 4: Similar right triangles at K and L give equal angles and a side ratio

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In plain words

The two perpendicular bisectors meet the angle bisector at the same angle $90^\circ - \dfrac{C}{2}$ , making the two right triangles $CKP$ and $CLQ$ similar.

\triangle CKP \sim \triangle CLQ \implies \frac{PK}{QL} = \frac{CP}{CQ}, \qquad \angle RPK = 90^\circ + \frac{C}{2} = \angle RQL

Detailed analysis

Let $C = \angle BCA$ , so $\angle KCP = \angle LCQ = \dfrac{C}{2}$ . Since $PK \perp BC$ at $K$ and $QL \perp AC$ at $L$ , right triangles $CKP$ and $CLQ$ are similar, giving $\dfrac{PK}{QL} = \dfrac{CP}{CQ}$ and $\angle CPK = \angle CQL = 90^\circ - \dfrac{C}{2}$ . Their supplements along line $CR$ therefore satisfy $\angle RPK = \angle RQL = 90^\circ + \dfrac{C}{2}$ .