Answer to Question #100400 in Geometry for Olivia

Question

Prove that in a right triangle, the bisector of the right angle halves the angle between the median and altitude drawn from the same vertex.

Answer 1

Let $ABC$ is the given triangle, $\angle ACB=90\degree,$ point $M,S\ and\ H$ belong $AB$, such that $CM$ is median, $CS$ is the bisector of $\angle ACB$, $CH$  is altitude.

In right triangle median to the hypotenuse is equal to half the hypotenuse, hence $|CM|=|AM|$ and triangle $AMC$ is isosceles, thus $\angle ACM=\angle BAC$.

Since $\angle CHB=90\degree,$ then $\angle HCB=90\degree-\angle ABC=90\degree-(90\degree-\angle BAC)=\angle BAC$ .

Since $CS$ is bisector, then $\angle SCA=\angle SCB=90\degree/2=45\degree.$

$\angle MCS=\angle SCA-\angle ACM=45\degree-\angle BAC$,

$\angle HCS=\angle SCB-\angle HCB=45\degree-\angle BAC$,

Thus $\angle MCS=\angle HCS$ and $CS$ halves the angle between the median $CM$ and altitude $CH$.