Solving for $DE$ in a Geometric Puzzle

Question

I hope this message finds you well. I am contacting you to seek your help in solving a fascinating geometry problem that I encountered in a recent competition. Despite my diligent attempts, I have not been able to find a solution. I am keen to acquire insights that will undoubtedly enhance my comprehension of this geometric puzzle.

Problem Description

In $\Delta$ $ACD$, $AC=AD=CD$, and $AB$ = $3$, $BC$ = $6$, find the value of $DE$

Approach: I tried this way to find the values of $x$ and $y$. therefore, I can apply the sine law.

I made a note of this method, and then I tried to determine the values of $x$ & $y$ but I was unable to do so... I would be extremely grateful for any help or advice in figuring out the intricacies of this problem.

Thank you for your expertise and support.

Answer 1

Hint 1: $AE:EC=1:2$

Consider the ratio of area $\Delta ABE:\Delta CBE$. They are triangle with common vertex, so the ratio is equal to the base ratio $AE:EC$. On the other hand we have the area of the triangles are $$\Delta ABE=\dfrac{1}{2}(3)(BE)\sin\angle ABE$$ $$\Delta CBE=\dfrac{1}{2}(6)(BE)\sin\angle CBE$$ As you have found, the angle are the same so the area ratio is $1:2$.

Hint 2: What is $AC$?

This is direct followed from cosine formula on $\Delta ABC$, where you can get $AC=\sqrt{63}$. And therefore, you get $AE,AD$ and $\angle DAE$. You get $DE$ as desired!