Prove that two lines are perpendicular in isosceles triangle geometrically We have given isosceles triangle $ABC$ with baseline $AB$. Point $M$ is midpoint of AB. Draw perpendicular line to side $AC$ through $M$ which intersects $AC$ in point $H$. Let $P$ be midpoint of $MH$. Show that lines $BH$ and $CP$ are perpendicular to each other.
I also have "hint":
Construct height $BD$ to side $AC$ and look at the triangles $ABD$ and $MCH$.
Analytically there is no problem but I get stuck in geometrical approach to solution.
 A: We draw the line $BD$ perpendicular to $AC$ and parallel to $MH$. Because $\angle CBA=\angle CAB=\angle CMH$, we know that $\triangle AMH\sim\triangle ABD\sim\triangle MCH$. Because $|AM|=|MB|$, it follows that $|AH|=|HD|$. Thus, $H$ is the midpoint of $AD$. Now, we see that $BH$ and $CP$ are medians in $\triangle ABD$ and $\triangle MCH$. Also, we can get $\triangle MCH$ by rotating $\triangle ABD$ over $90^\circ$ and translating and scaling it. Because translation and scaling don't change the angle between corresponding lines, we now find that $BH\perp CP$.

A: I hope the following can satisfy the requirement of a geometrical solution.
I modified Ragnar’s drawing as 
As pointed out:-
(1) HD = (1/2) AD ------------------------ [intercept theorem]
(2) △ABD ∼ △MCH-------------------- [equi-angular]
Then, from (2), AD / BD = MH / CH ----------- (*)
In △HCP, tan y = HP / CH 
Then, tan y = (1/2)(MH) / CH -------------------- [given]
In △HBD, tan z = HD / BD 
Then tan z = (1/2)(AD) / BD --------------------- [from (1)]
From (*), we have tan y = tan z 
Hence, y = z [since none of the angles are more than π.]
Then, C, D, K and B are con-cyclic ---------[converse of angles in the same segment]
angle CKB = angle CDB = 90 degree------ [angles in the same segment]
