Given that $\triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=\frac{1}{2}BD$, prove that BD bisects angle $\angle ABC$.
I have tried proving triangle $\triangle AEB$ and triangle $\triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $\angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.
Best Answer
A simple geometric solution:
Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=\frac {1}{2}DB=\frac {1}{2}AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.