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Belarusian Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Belarus

Let $P$ be the intersection point of the diagonals of inscribed quadrilateral $ABCD$ . Points $K$ and $L$ are marked on the bisectors of $\angle APD$ and $\angle BPC$ , respectively, so that $AP = PK$ and $BP = PL$ . Let $M$ be the intersection point of the lines $AK$ and $BL$ , and $N$ be the intersection point of the lines $KD$ and $LC$ .
Prove that the lines $KL$ and $MN$ are perpendicular.
(D. Pirshtuk)

Solution

Let $\angle APD = \angle BPC = \varphi$ . Then $\angle PKA = \angle PLB = \frac{\pi-\varphi}{2}$ since the triangles $APK$ and $BPL$ are isosceles. Therefore, $MK = ML$ .

From the equalities $AP = PK$ , $BP = PL$ and the power of a point theorem it follows that $PK \cdot PC = AP \cdot PC = BP \cdot PD = PL \cdot PD$ . Hence, $\frac{PK}{PL} = \frac{PD}{PC}$ . Further, $\angle KPD = \angle LPC = \frac{\varphi}{2}$ . Therefore, the triangles $KPD$ , $LPC$ are similar, so $\angle PKD = \angle PLC$ . Hence, $NK = NL$ .
Now we have $MK = ML$ and $NK = NL$ . It follows that $M$ and $N$ lie on the perpendicular bisector of the segment $KL$ , which gives the required statement.

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