Geometry problem involving median and bisector

Question

I'm not being able to solve this geometry problem. So I thought you could help me.

CR is a bisector. AM is a median. BÂC is 90º. Find x.

Answer 1

There's a right angle at the top, so this is a circle with M at the centre of the circle.

Angle BMA is 100 and triangle BMA is isosceles so angles MBA and MAB are both 40.

Angle ACB + CBA = 90 so ACB = 50. So RCB must be 25 and x is 65.

Sorry if this is unclear, it's difficult to explain with words. If anything is confusing, please ask.

Answer 2

Since $BAC$ is a right triangle we have $MB=MC=MA$. $\widehat{HMA}=80^\circ$ hence $\widehat{BMA}=100^\circ$ and $\widehat{MBA}=\widehat{CBA}=40^\circ$. From $\widehat{BCA}=50^\circ$ it follows that $\widehat{RCB}=25^\circ$ and $x=\color{red}{65^\circ}$.