In triangle $ABC$, prove that the angle between the bisector of $A$ (call $AD$) and the height $AH$ is: $$\measuredangle HAD=\frac{|\measuredangle B-\measuredangle C|}{2}$$ The book states that by writing the equations of different angles, problem could be easily deducted, but I tried and didn't get anywhere.

$$\measuredangle HAD = \measuredangle HAC - \measuredangle DAC = 90^{\circ} - \measuredangle C - \frac{\measuredangle A}{2} =$$ $$=\frac{\measuredangle A}{2} + \frac{\measuredangle B}{2} + \frac{\measuredangle C}{2} - \measuredangle C - \frac{\measuredangle A}{2} = \frac{\measuredangle B}{2} - \frac{\measuredangle C}{2}.$$

(I have assumed angle biscetor to be nearer than AH near AC).

The interesting fact from here you get that if $O$ is the circumcenter of $\Delta ABC$ , then $\measuredangle\ HAC = \measuredangle BAO =90^{\circ} -\measuredangle C$.

Hence the $AH$ reflected with resepct to the angle bisector gives $AO$.

Thus $AH$ and $AO$ are said to be $\textbf{isogonal}$ with respect to $\angle A$.

Thus similarly with $\angle B$ and also $\angle C$.

Thus $H$ and $O$ are said to be $\textbf{isogonal conjugate}$ with respect to triangle $ABC$