Source: Regional Math Olympiad of BD.
In the given diagram, both $\angle ABC$ and $\angle DBE$ are right-angled triangles. $B$ is the right angle for both (according to the diagram). They have hypotenuses $AC$ and $DE$ of the same length and $F$ is the middle point of both the hypotenuse. $DE$ is perpendicular to $BC$. Area of $\triangle ABC$ is $\sqrt 3$ times of $DEF$. What is the value of angle $\angle BDE$?
Here is a likelier figure of the contextual problem:
At first, I thought $DEF$ meant $DE+EF$. Therefore, $DE$ + $\frac{1}{2} DE$ = $\frac {3}{2} DE$. As it is mentioned in the above post that $AC= DE$, so $\frac {3}{2} DE$ = $\frac{3}{2} AC$.
Let us denote the angle $\angle ACB$ = $\theta$
From another condition, we can write that:
$\frac{1}{2} AB × BC$ = $\frac {3 \sqrt 3}{2} AC$.
$AB×BC$ = $3 \sqrt 3 AC$
And then, dividing with $BC$ to both side, we get:
$AB$ = $\frac {3 \sqrt 3 C}{BC}$
$AB$ = $3 \sqrt 3 \sec\theta$ (by denoting $\angle ACB$ as $\theta$)
Here is the problem. I failed to show any relation of $\theta$ with $AB$. I don't know whether I'm approaching towards the right direction or not. If not so, then please help me find my mistake.
Kindly pardon my error. Thanks in advance.
And if you correct my mistake, then I'll very much glad.
Best Answer
I replace the alleged $\triangle DEF$ by $\triangle DEB$.
Let $|FG|=:h$, $|BG|=:s$, and let $r$ be the radius of the circle with center $F$ going through $A$, $B$, E$, C$, and $D$. Then ${\rm area}(ABC)=\sqrt{3}\>{\rm area}(DEB)$ implies $2hs=\sqrt{3}rs$, so that together with $h^2+s^2=r^2$ we obtain $$\angle(BFG)=\arctan{s\over h}={1\over\sqrt{3}}=30^\circ\ .$$ This implies $\angle(BDG)=15^\circ$.