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Problems of Ukrainian Authors

Geometry Difficulty 6.3 National Olympiad Prove it Ukraine

Given $ABCD$ be a quadrilateral inscribed in a circle with center $O$ , and let $P = AC \cap BD$ , $BC \parallel AD$ . Rays $AB$ and $DC$ intersect at the point $E$ . A circle with the center $I$ is inscribed in the triangle $EBC$ and is tangent to the line $BC$ at point $T_1$ . The excircle of the triangle $EAD$ with center $J$ is tangent to the side $AD$ at point $T_2$ . The lines $IT_1$ and $JT_2$ intersect at the point $Q$ . Prove that points $O$ , $P$ and $Q$ are collinear.

Solution

Let $P_1$ , $P_2$ be respectively the feet of perpendiculars from $P$ to lines $BC$ and $AD$ , $M_1$ , $M_2$ respectively are midpoints of sides $BC$ and $AD$ respectively, the excircle of the triangle $EBC$ touches the side $BC$ at the point $T'$ . $T'C = BT_1 \Rightarrow \frac{BT_1}{T_1C} = \frac{CT'}{T'B}$ . $ABCD$ is a cyclic quadrilateral then $\triangle EBC \sim \triangle EAD \Rightarrow \frac{BT_1}{T_1C} = \frac{CT'}{T'B} = \frac{AT_2}{T_2D}$ (Fig.22).

Moreover $\triangle PBC \sim \triangle PAD \Rightarrow \frac{BP}{P_1C} = \frac{AP_2}{P_2D}$ . Since $M_1$ , $M_2$ are the midpoints of segments $BC$ and $AD$ respectively then $\frac{T_1M_1}{M_1P_1} = \frac{T_2M_2}{M_2P_2}$ .

Let the point $O_1$ be located on the segment $PQ$ such that $\frac{QO_1}{O_1P} = \frac{T_1M_1}{M_1P_1}$ . Then the projections of this point on the segments $BC$ and $AD$ respectively coincide with points $M_1$ and $M_2 \Rightarrow O$ coincides with $O_1$ . Then points $O$ , $P$ and $Q$ are collinear.

Fig.22

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