China Western Mathematical Olympiad

Let $A_1$ be the center of A-excircle. By the properties about centers of ex-circles, the following sets of three points are collinear: $\{B_1,A,C_1\}$, $\{A_1,C,B_1\}$ and $\{C_1,B,A_1\}$, and $A_{1}A\perp B_1{C}_1$.

Fig. 3. 1

On the plane, choose point $P$ such that $\overrightarrow{C_{2}P}=\overrightarrow{B_{2}C}$, then it follows from $\overrightarrow{B_{2}C}=\overrightarrow{A{B}_1}$ that $\overrightarrow{C_{2}P}=\overrightarrow{A{B}_1}$. As $\overrightarrow{B{C}_2}=\overrightarrow{C_{1}A}$ and the points $C_1,B,A_1$ are collinear, the points $B,C_2,P$ are collinear, so $\overrightarrow{BP}=\overrightarrow{C_1{B}_1}$. It follows from $$\begin{array}{rl}\angle A{C}_{1}B& =180^{\circ }-\left(\frac{180^{\circ }-\angle BAC}{2}\right)-\left(\frac{180^{\circ }-\angle ABC}{2}\right)\\ & =\frac{\angle BAC+\angle ABC}{2}=\frac{180^{\circ }-\angle ACB}{2}=\angle BC{A}_1\end{array}$$ that $△A_{1}BC\sim △A_1{B}_1{C}_1$. Let $A_{1}D$ and $A_{1}A$ be the altitudes of the $△A_{1}BC$ and $△A_1{B}_1{C}_1$, respectively, with respect to the opposite sides, so $\frac{B_1{C}_1}{BC}=\frac{A_{1}A}{A_{1}D}$.

If $\overrightarrow{BP}=\overrightarrow{C_1{B}_1}$, then $A_{1}A\perp B_1{C}_1$, so $\frac{BP}{BC}=\frac{A_{1}A}{A_{1}D}$. If $BP\perp A_{1}A$, then $BC\perp A_{1}D$, so we have $△BPC\sim △A_{1}AD$, and hence $CP\perp AD$. Again by $\overrightarrow{C_{2}P}=\overrightarrow{B_{2}C}$, one has $\overrightarrow{B_2{C}_2}=\overrightarrow{CP}$, and hence $AD\perp B_2{C}_2$.