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 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 $\hat B > \hat C$. What does that leave you with?
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The altitude drawn to the hypotenuse of a right triangle forms two congruent triangles
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