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Mongolian National Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Mongolia

Let $\omega$ be circumcircle of triangle $ABC$ and let $AD$ and $BE$ be altitudes. A line $DE$ intersects the circle $\omega$ at points $P$ and $Q$ with order $P$ , $E$ , $D$ , $Q$ in the line. Let bisectors of angle $APQ$ and $BQP$ intersect a circle $\omega$ at point $K$ and $L$ , respectively. Prove that the line $KL$ is perpendicular to the bisector of angle $ACB$ .
(Proposed by B. Ulziinasan)

Solution

Let $O$ be a circumcenter of triangle $ABC$ . We know $\angle BCO = 90^\circ - \angle A$ and $\angle CDE = \angle A$ so $\angle BCO + \angle CDE = (90^\circ - \angle A) + \angle A = 90^\circ$ , from here $\text{PQ CO}$ .

Hence $\overline{CQ} = \overline{CP}$ and denote it by $x$ . If we denote $\overline{QB} = 2y$ , $\overline{BA} = 2z$ , $\overline{AP} = 2t$ then $\overline{BM} = \overline{MA} = z$ . Because of $\overline{QK} = \overline{KA}$ we have $\overline{QA} = 2(y+z)$ and $\overline{QK} = y+z$ . From here $\overline{KM} = y$ .
Because $\overline{BL} = \overline{LP}$ , we have $\overline{BP} = 2(t+z)$ and $\overline{LP} = t+z$ . So we get $\overline{LC} = z+t+x$ . Now we can write $\overline{KM} + \overline{LC} = y + z + t + x = \overline{CQ} + \overline{QB} + \overline{BK} + \overline{KM} = 180^\circ$ , in other words $\overline{KL} \perp \overline{MC}$ .

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