As shown in the figure, in the parallelogram $ABCD$, $I$ is the incenter of $\vartriangle BCD$, and $H$ is the orthocenter of $\vartriangle IBD$. Prove that $\angle HAB=\angle HAD$.
Source: China Northern MO 2014 p5 CNMO
Tags: geometry, equal angles, parallelogram
As shown in the figure, in the parallelogram $ABCD$, $I$ is the incenter of $\vartriangle BCD$, and $H$ is the orthocenter of $\vartriangle IBD$. Prove that $\angle HAB=\angle HAD$.