In right $\Delta ABC$, $\angle C = 90^\circ$. $E$ is on $BC$ such that $AC = BE$. $D$ is on $AB$ such that $DE \perp BC$ . Given that $DE + BC = 1$ and $BD = \frac{1}{2}$, find $\angle B$.
What I Tried: Here is a picture :-
The first try which I think I could do was the Pythagorean Theorem, although I am not sure whether that will help me find $\angle B$. Let $DE = x$, then we have :- $$BE = \sqrt{\frac{1}{4} – x^2}.$$
Now as $AC = BE$, $AC = \sqrt{\frac{1}{4} – x^2}$ too.
Now, let $CE = y$. What I can do is find $AD$ from similar triangles $BED$ and $BAC$ as well as from Pythagorean Theorem, and then I suppose find the variables $x$ and $y$ but in the end I suppose I am not finding $\angle B$. I also didn't try all this because I think it won't work and is going to be complicated.
Can anyone solve this? I didn't try using Trigonometry because I am a little weak at it, so any solutions without Trigonometry will be appreciated.
Best Answer
Let $DE=x$, $AC=BE=y$.
$$ \dfrac{DE}{BE} = \dfrac{AC}{BC}$$
$$ \dfrac{x}{y} = \dfrac{y}{1-x}$$
$$ \Rightarrow x = x^2+y^2 = \dfrac{1}{4}$$
$$ \sin B = \dfrac{DE}{BD}=\dfrac{1}{2}$$
$$\therefore \angle B = 30^{\circ}$$