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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 $\angle{ABC}$.

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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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    $\begingroup$ Welcome to MSE! Just a few tips: 1. Whenever you're typing an equation, make sure to use dollar signs before and after the expression to make sure it is rendered. 2. It would be wise to show your attempt at the question. Questions of the form: 'Please solve this for me' are usually poorly received. $\endgroup$
    – Vector
    Commented Feb 20 at 6:59
  • $\begingroup$ How to show my attempts regarding this problem; and how to add diagram in it; please help I have joined newly. $\endgroup$
    – Mathematical Physicist
    Commented Feb 20 at 7:10
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    $\begingroup$ You can construct your answer the way you did your question. Just remember to use dollar signs, $ to ensure the equation is rendered. $\endgroup$
    – Vector
    Commented Feb 20 at 7:11
  • $\begingroup$ how to add diagram in it $\endgroup$
    – Mathematical Physicist
    Commented Feb 20 at 7:35
  • $\begingroup$ @MathematicalPhysicist: for geometry questions, it's always a good idea to add a drawing of the situation. Plenty of tools on the internet for creating such images (personally I use Geogebra). $\endgroup$
    – Dominique
    Commented Feb 20 at 7:38

2 Answers 2

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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 $A\hat{B}C$.

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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 $\alpha = \beta$

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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 $\alpha=\beta$.

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