# Congruency and congruent triangles

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 ????

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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 $$A\hat{B}C$$.

Here's a simple solution. The original information is in black, where I've marked the sides of the isosceles as $$s$$ and the length of $$AE$$ as $$x$$. Extend $$AE$$ to meet $$BC$$ at $$F$$ (shown in red), mark the corresponding angles of $$\alpha$$, and also mark the angle $$\beta$$ where we seek to prove $$\alpha = \beta$$

Now, $$(x+z)\cos(\alpha) = s = 2x \cos(\alpha) \;\Rightarrow\; z=x$$

Thus $$AF$$ is bisected at $$E$$, so $$\triangle FEB \cong \triangle AEB$$, and so $$\alpha=\beta$$.