Prove that the midpoint lies on the right angle bisector. 2 equal circles are located inside a right-angled triangle so that they touch each other and each circle also touches one leg and the hypotenuse. Let M and N be points of tangency of the 2 circles with the hypotenuse respectively. Provide a proof that shows that the midpoint of MN lies on the right angle bisector of the given triangle.
I have drawn the diagram and have been experimenting with dividing the inscribed circle into 2 parts and making the triangle isosceles however, I am still confused about how to provide a proof for it. Help would be appreciated. 
Thank you. 
 A: 
Here $\Delta ABC$ is the right-angled triangle and $R$ is the midpoint of $MN$. Let the common radius of the two circles be $r$. Let $\angle MAO_1=\theta$
Making use of the Angle Bisector Theorem, to prove that $BR$ is the angle bisector of $\angle ABC$, it will be sufficient to show that $\dfrac{AR}{RC}=\dfrac{AB}{BC}$
Simple trig tells you that $\color{blue}{AM=r\cot\theta}$ and $\color{blue}{CN=r\cot(45^\circ-\theta)}$. Since $O_1MNO_2$ is a rectangle, $\color{blue}{MR=RN=r}$. This makes $\color{blue}{AR=r(1+\cot\theta)}$ and $\color{blue}{RC=r(1+\cot(45^\circ-\theta))}$
So the ratio becomes: $$\dfrac{AR}{RC}=\dfrac{1+\cot\theta}{1+\cot(45^\circ-\theta)}=\dfrac{1+\cot\theta}{1+\dfrac{\cot\theta+1}{\cot\theta-1}}=\dfrac{(1+\cot\theta)(\cot\theta-1)}{2\cot\theta}=\dfrac{\cot^2\theta-1}{2\cot\theta}=\cot2\theta$$
Simple trig also tells you that $\dfrac{AB}{BC}=\cot2\theta$. This completes the proof.
A: The claim can not be right - the midpoint of the hypotenuse (based on the conditions) lies on the bisector only if the right angle triangle is an isosceles one.
