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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?

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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.

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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. $H \in PA'$. enter image description here

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 $BC$ $$AH = A'H = BH$$

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