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Geometry Difficulty 6.2 National Olympiad Prove it Taiwan

Let $A$ , $B$ , $C$ be the midpoints of sides $B'C'$ , $C'A'$ , $A'B'$ of $\triangle A'B'C'$ , respectively. Let point $P$ lie in the interior of $\triangle ABC$ , and let $AP$ , $BP$ , $CP$ meet $BC$ , $CA$ , $AB$ at points $P_a$ , $P_b$ , $P_c$ , respectively.
Lines $P_aP_b$ , $P_aP_c$ meet $B'C'$ at $R_b$ , $R_c$ respectively; lines $P_bP_c$ , $P_bP_a$ meet $C'A'$ at $S_c$ , $S_a$ respectively; lines $P_cP_a$ , $P_cP_b$ meet $A'B'$ at $T_a$ , $T_b$ respectively. It is known that $S_c$ , $S_a$ , $T_a$ , $T_b$ all lie on a circle with center $O$ .
Prove: $OR_b = OR_c$ .

Solution

First we prove that $A$ is the midpoint of $R_bR_c$ : since $BC \parallel B'C'$ , we know $\frac{AR_c}{BP_a} = \frac{AP_c}{P_cB}$ and $\frac{AR_b}{CP_a} = \frac{AP_b}{P_bC}$ . From these two equations and Ceva's theorem, we know $\frac{AR_c}{AR_b} = \frac{AP_c}{P_cB} \cdot \frac{BP_a}{P_aC} \cdot \frac{CP_b}{P_bA} = 1$ . Clearly $R_b$ , $R_c$ lie on opposite sides of $A$ , so $A$ is the midpoint of $R_bR_c$ .

Similarly, we can prove that $B$ is the midpoint of $S_cS_a$ , and $C$ is the midpoint of $T_aT_b$ . Therefore $OB \perp C'A'$ , $OC \perp A'B'$ . Hence $O$ is the circumcenter of $\triangle A'B'C'$ (or the orthocenter of $\triangle ABC$ ). We obtain $OA \perp B'C'$ , and hence $OR_b = OR_c$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.