In an isosceles right $\Delta ABC$, $\angle B = 90^\circ$. AD is the median on BC. Let $AB = BC = a$. If $BE \perp AD$, intersecting $AC$ at $E$, and $EF \perp BC$ at $F$, find $EF$ in terms of $a$.
What I Tried: Here is a picture,
I saw a lot of similar triangles here, because the coloured angles are equal., but I couldn't find any use of them in my case.
I also used Pythagorean Theorem in a lot of ways. First we have $AB = BC = a$ , so :- $$AC = \sqrt{2}a$$ Also, $BD = DC = \frac{a}{2}$ , so :- $$AD = \frac{\sqrt{5}a}{2}$$. After that I assumed $AG$ to be $x$ and $GD$ to be $(\frac{\sqrt{5}a}{2} - x)$, and $BG = y$.
From here I got some more information, such as :- $$(i) \sqrt{5}x = 2a$$ $$(ii) x = 2y$$
I didn't proceed further, as I didn't know what to denote the side $AE$ and E$C$ with, and after putting another variable it gets over-complicated. I am stuck here.
Can anyone help? (Solution without Trigonometry will be better) .
The two pairs of similar triangles you need are $\triangle ABC \sim \triangle EFC$ and $\triangle ABD \sim \triangle BFE$.
Using them, we have the equations:
$$\frac {AB}{BC} = \frac {EF}{FC} = 1, \quad \frac{AB}{BD} = \frac{BF}{FE} = 2 = \frac {BF}{FC}$$
Now we can find $EF = FC$ in terms of $a$.