Given that $\Delta\,ABC$ is an isosceles right triangle with $AC = BC$ and $\angle{ACB}= 90^\circ$. D is a point on AC and E is on the extension of BD such that $AE \perp BE$. If $AE = \frac{1}{2}BD$, prove that BD bisects $\angle{ABC}$.
Since $AE = \frac{1}{2}BD$, so I mark point F on BD such that it is the midpoint of BD and then draw FG i.e G lies on AB and FG = AE. By using some angle chasing I find $\angle{DGA}$ a right angle. From here my aim is to make $\Delta\,DCB$ congruent to $\Delta DGB$ but I cannot able to deduce the solution from here please help me ????
Let $F$ be the midpoint of $BD$, then draw lines $CF$ and $CE$.
Quadrilateral $ABCE$ is cyclic because $A\hat{E}B = A\hat{C}B$, so angles $C\hat{E}B$ and $C\hat{A}B$ are equal since they determined by the same arc. Knowing $C\hat{A}B = 45^{\circ}$, then $C\hat{E}B$ is also equal to $45^{\circ}$ and angle $C\hat{E}A = 135^{\circ}$ since it's equal to $C\hat{E}B + A\hat{E}B$.
Since $F$ is the midpoint of the hypotenuse of $\triangle BCD$, it is the circumcenter of the triangle, which means it is equidistant from all vertices, thus $CF = BF = DF$, and since $BF = \dfrac{BD}{2} = AE$, we also know $CF = AE$.
Note that $E\hat{B}C = E\hat{A}C$, since they are determined by the same arc. But then triangles $\triangle BCF$ and $\triangle ACE$ are congruent, since two pairs of corresponding sides and the corresponding angles between them are equal. Now, we have $CE = CF = AE$, which means triangle $\triangle ACE$ is isosceles and $E\hat{A}C = A\hat{C}E = \dfrac{180^{\circ} - 135^{\circ}}{2} = \dfrac{45^{\circ}}{2}$. But we already know $E\hat{A}C = E\hat{B}C$, so $E\hat{B}C = \dfrac{45^{\circ}}{2}$, which means $BE$ bisects $A\hat{B}C$.