How would I go about proving this:
In a right triangle, the median and the altitude drawn to the hypotenuse make an angle congruent to the difference of the acute angles of the triangle.
One solution is to construct each angle with a compass, subtract them, and then prove congruence through superimposing one on the other – but I'm sure there must be a more theoretical approach.
Best Answer
Let $BC$ be the hypotenuse, $AH$ the altitude corresponding to $BC$ and $AM$ the median that corresponds to $BC$.
Due to a theorem, we have that $AM=\dfrac{BC}2=BM=MC$. Thus, we have 2 isosceles triangles $\triangle ABM, \triangle AMC$.
So, we have that $$\angle ABM=\angle BAM=x \text{ and } \angle MAC =\angle ACM=y.$$
Moreover, $\angle HMA=2y$ as exterior angle of $\triangle AMC$.
Working in the right triangle $\triangle HMA$, we have that the blue angle is: $$\angle HAM=90^\circ-2y\, (1)$$
However, $$x+y=90^\circ\, (2)$$
From (1), (2) we have that the blue angle is: $$\angle HAM =90^\circ-2y=x+y-2y=x-y.$$
Note: Without loss of generality, we considered $\hat B > \hat C$.