Find the radius of the inscribed circle of a right triangle. The triangle's height is $\sqrt{6} + \sqrt{2}$ while the bisector of the right angle is 4.
This seemed like a generic similar triangles problem but it isn't that simple when I tried to solve it. Any help?
EDIT: Here is a picture of the problem
Best Answer
Let $s=|CA|+|BC|$ and $p=|CA|\cdot|BC|$ then the double area of the triangle can be written as $$p=|CD|\cdot|AB|=|CE|s\sin(\pi/4).$$ Now in order to find the inradius $r$ recall that $$r=\frac{p}{s+|AB|}=\frac{p}{\frac{\sqrt{2}p}{|CE|}+\frac{p}{|CD|}}=\frac{1}{\frac{\sqrt{2}}{|CE|}+\frac{1}{|CD|}}.$$ Can you take it from here?