PROBLEM
Let $ABC$ be a right triangle at $A$. The perpendicular at $C$ on $BC$ and the median of side $AC$ intersect at $S$. Show that $\angle SBA=\angle MBC$, where $M$ is the midpoint of side $AC$.
WHAT I THOUGHT OF:
DRAWING
link to the drawing and the auxiliar construction
Okey, we can simply demonstrate that $MS\parallel AN$.
Because of the parallelism and the fact that $M$ is the midpoint of $AC$, applying the the reciprocal of the middle line theorem we got that $S$ is also the midpoint of $NC$.
I saw that $\bigtriangleup ASN $ and $\bigtriangleup BMC$ look like similar triangles. They have $\angle BCA$ and $\angle BNC$ congruence.
I dont know how to provide the next information to show the similarity.
I thought of showing that $\frac{BS}{BM}=\frac{NS}{CM}$. $BS$ and $BM$ are both median in some triangles so we can apply the median formula. I tried applying it, but i dont succed showing what i want to show.
Hope one of you can help me! Thank you!
You are quite close.
$$\triangle ABC \sim \triangle CBN \Rightarrow \angle C = \angle N$$
Also $$\frac{AC}{BC}=\frac{CN}{BN} \Rightarrow \frac{AC/2}{BC}=\frac{CN/2}{BN} \Rightarrow\frac{MC}{BC}=\frac{SN}{BN} $$
$$\therefore \quad \triangle BMC \sim \triangle BSN \Rightarrow \angle SBA=\angle MBC \quad \square$$