In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AB$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.
parmenides51 wrote:
In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AC$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.
To respect symmetry, $C_1$ must be the midpoint of $AB$. In that case let $P'$ be the reflection of $B$ in $P$. We have $$\angle PP'A = \angle BP {{C}_{1}} = \angle PCA$$Hence $AP'CP$ is cyclic and we will have
$$\angle BP {{A}_{1}}= \angle PP'C = \angle PAC$$