# Prove that $BH=AH$

A triangle $$ABC$$ is given. There's a point $$P$$ inside it and also it is connected to point $$H$$, which lies on edge $$BC$$ ($$H$$ must not be the middle point of edge $$BC$$). Turns out, that bisector of angle $$∠AHP$$ is perpendicular to edge $$BC$$. Also, $$BP=AC$$ and $$∠PCH=∠ABC$$. Prove that $$BH=AH$$.
What I found out was that since triangle $$ACH$$ hypothetically should be equal to triangle to $$PBH$$, we could prove what we need by proving that for example $$PH=CH$$ - though again, I have no idea how to do it. Could you recommend me any smart lines or segments to draw? Since the angle bisector of $$\angle AHP$$ is perpendicular to $$BC$$, $$\angle PHC = \angle AHB$$. This and $$\angle PCH = \angle HBA$$ imply that $$\triangle PHC$$ and $$\triangle AHB$$ are similar. Thus $$\frac{PH}{CH} = \frac{AH}{BH},$$ or $$PH\cdot BH = AH \cdot CH.$$ From this and $$BP=AC, \angle BHP=\angle CHA$$, the cosine law in $$\triangle PBH$$ and $$\triangle CAH$$ would give $$PH^2 + BH^2 = AH^2 + CH^2.$$
Thus $$(PH,BH) = (AH, CH)$$ or $$(PH,BH) = (CH,AH)$$. But $$H$$ is not the midpoint of $$BC$$, i.e. $$BH\neq CH$$, so we are done.
Since the angle bisector of $$\angle \, AHP$$ is perpendicular to $$BC$$, then $$BC$$ is in fact the outer angle bisector of $$\angle \, AHP$$ so $$\angle \, AHC = \angle \, PHB$$.
Let point $$A'$$ be the reflection image of point $$A$$ with respect to the line $$BC$$. Then triangle $$\Delta \, A'BC$$ is the reflection image of triangle $$\Delta\, ABC$$ with respect to $$BC$$. Consequently, $$\angle \, A'HC = \angle \, AHC = \angle\, PHB$$ which is possible if and only if the points $$P, H$$ and $$A'$$ are collinear, i.e. $$H \in PA'$$. By assumption, $$\angle \, PCB = \angle \, BCB = \angle \, ABC$$ But by construction, $$\angle \, ABC = \angle \, A'BC$$ which yields $$\angle \, PCB = \angle \, A'BC$$ and therefore the segments $$CP \, || \, A'B$$ i.e. the quad $$CPBA'$$ is a trapezoid. By assumption, however, $$BP = AC$$ and by construction (reflection symmetry) $$AC = A'C$$ so $$BP = AC$$ Therefore the trapezoid $$CPBA'$$ is isosceles, which implies that $$A'H = BH$$. But since $$A'$$ is the symmetric image of $$A$$ with respect to $$BC$$ $$AH = A'H = BH$$