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 4 of 4: Compute the area ratio and conclude equality

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In plain words

Equal included angles cancel the sine factors, and the two ratios $\dfrac{RP}{RQ} = \dfrac{CQ}{CP}$ and $\dfrac{PK}{QL} = \dfrac{CP}{CQ}$ are exact reciprocals.

\frac{[\triangle RPK]}{[\triangle RQL]} = \frac{\frac{1}{2} RP \cdot PK \sin \angle RPK}{\frac{1}{2} RQ \cdot QL \sin \angle RQL} = \frac{RP}{RQ} \cdot \frac{PK}{QL} = \frac{CQ}{CP} \cdot \frac{CP}{CQ} = 1

Detailed analysis

Using $\angle RPK = \angle RQL$ (Step 1), $\dfrac{PK}{QL} = \dfrac{CP}{CQ}$ (Step 1), and $\dfrac{RP}{RQ} = \dfrac{CQ}{CP}$ (Step 3), the ratio of areas is $\dfrac{[\triangle RPK]}{[\triangle RQL]} = \dfrac{RP \cdot PK}{RQ \cdot QL} = \dfrac{CQ}{CP} \cdot \dfrac{CP}{CQ} = 1$ , so $[\triangle RPK] = [\triangle RQL]$ .