Perpendicular lines in Isoceles triangle So I was reviewing my olimpiad materials for Geometry and in the notes I took in different classes I found this question:
Consider Isoceles triangle $ABC$ $(AB=AC)$ and let $M$ be the midpoint of $BC$. We draw $MH$ perpendicular to $AC$. Now let $N$ be the midpoint of $MH$, prove that $AN$ and $BH$ are perpendicular.
Any ideas for a solution!? I tried finding a cyclic quadrilateral and drew some parallels with MH to maybe create some helpful similarities but got to no where!
 A: 
Hints:
Join B to H. Draw a perpendicular from M on BH, it intersect the extension of CA at E. Points M, H and E are on a circle center at O, because $\angle MHE=90^o$. Join A to O and extend it to meet the circle at P. Join P to G. PG is the diameter of the circle(explain why). In this way $\angle PAG=90^o$ so $OA||MH$. OA bisects EH , that is A is midpoint of EH. N is the midpoint of MH that means $AN||ME$ ,therefore $AN\perp BH$.
A: 
Draw $BK \perp AC$ and note that in $\triangle BKC$ , $KH = HC$ (why?). Therefore in this triangle $BH$ is the median of $KC$.
Now, note that $\triangle BKC$ and $\triangle AHM$ are similar. In fact one can obtain one of these triangles from the other, by proper scaling, displacement, and a $90^o$ rotation (why?).
Can you take it from here?
A: One can show that the triangles $\Delta AMN$ and $\Delta BCH$ are similar.
Their corresponding angles in $M$ and $C$ are equal, having the same complement
$\widehat{MAC}$. We need one more proportionality.
(A possibility to work geometrically now is to draw the height $BB'$, which is double $MH$, and compare lengths of segments.) The trigonometric solution is simpler to type and follow, so i will go this way...
We may and do assume $AB=AC=1$.
Let $x$ be the measure of the angles $\widehat{BAM}=\widehat{MAC}$. Then we express the lengths of some segments in terms of $x$ as follows:
$$
\begin{aligned}
BM=MC&=\sin x\ ,\\
AM &=\cos x\ ,\\
HC &=\sin^2x\ ,\\
MH &=\sin x\cos x\ ,\\
MN=NH &=\frac 12\sin x\cos x\ ,
\end{aligned}
$$
and now we finish in a line the proportionality:
$$
\frac{AM}{MN}
=\frac{\cos x}{\frac 12\sin x\cos x}
=\frac{2}{\sin x}
=\frac{2\sin x}{\sin^2 x}
=\frac {BC}{CH}\ .
$$
So the angles in $A$ and $B$ of the mentioned (now known to be) similar triangles are equal. From the perpendicularity of sides $AM\perp BC$ we obtain then the perpendicularity of sides $A\perp BH$.
$\square$
