A geometry problem by Alan Enrique Ontiveros Salazar

Let the angle bisector be $s$ .

The area of the whole triangle is $\frac{1}{2}ab$ ; the area of the smaller triangles are $\frac{1}{2}as\sin \frac{\pi}{4}$ and $\frac{1}{2}bs\sin \frac{\pi}{4}$ .

It is then easy to see that

$\frac{1}{2}ab = \frac{1}{2}as\sin \frac{\pi}{4}+\frac{1}{2}bs\sin \frac{\pi}{4}$

After a bit of simplification:

$ab = \frac{1}{\sqrt{2}}as+\frac{1}{\sqrt{2}}bs = \frac{s(a+b)}{\sqrt{2}}$

Dividing both sides by $\frac{a+b}{\sqrt{2}}$ gives a formula for $s$ :

$s = \boxed{\frac{\sqrt{2}ab}{a+b}}$

Draw $ED$ parallel to $BC$ dividing $b$ so that $b=x+y$ , which makes $y=b-x$ .

From the similarity of $\triangle BCA$ and $\triangle EDA$ we have $\frac{a}{b}=\frac{x}{y}=\frac{x}{b-x}$ .

Solving for $x$ we get $x=\frac{ab}{a+b}$ .

$CE=\sqrt{2}x=\frac{\sqrt{2}ab}{a+b}$ .