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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 ????

Mathematical Physicist
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    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. – Vector Feb 20 '24 at 06:59
  • How to show my attempts regarding this problem; and how to add diagram in it; please help I have joined newly. – Mathematical Physicist Feb 20 '24 at 07:10
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    You can construct your answer the way you did your question. Just remember to use dollar signs, $ to ensure the equation is rendered. – Vector Feb 20 '24 at 07:11
  • how to add diagram in it – Mathematical Physicist Feb 20 '24 at 07:35
  • @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). – Dominique Feb 20 '24 at 07:38
  • @MathematicalPhysicist: I believe that a small reputation should be enough to modify your question and add an image. I've just upvoted your question. This gives you a reputation of 11. Can you add the image to your question now? – Dominique Feb 20 '24 at 07:59
  • @MathematicalPhysicist There will be an image icon where you type your answer. That allows you to upload an image from your device. – D S Feb 20 '24 at 08:29
  • I have edited my question. – Mathematical Physicist Feb 20 '24 at 13:58

2 Answers2

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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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I'm Ingrid
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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 $\alpha = \beta$

enter image description here

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 $\alpha=\beta$.

RobinSparrow
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