What's the measure of the angle $\angle DAC$ in the triangle below?

Question

For reference: In the figure, $AB = 11$, $BC = 5$ and $DE = 3$. Calculate $\angle DAC$.

My progress:

Extend $BE$ until$I: F\in AD$

Angle chasing: $HBI = \theta$

Can I say that the triangle $ABI$ is isosceles e therefore $\angle AIB = \theta \implies$

$\triangle ABI: 90^o = 3\theta \therefore \theta = 30^o$?

Answer 1

I don't see really why is $ABI$ isosceles.

Let $S$ be a midpoint of $AH$ and $x= CE = BE$ and $BH = t$.

If angle at $E$ is realy right then we have $$\triangle AEC\sim \triangle ABH \implies {16\over 2r} = {x+3\over t}$$ and $$\triangle BCD\sim \triangle BHS \implies {5\over t} = {x\over r}$$

We get $${t\over r} = {5\over x} = {x+3\over 8} \implies x^2+3x-40 =0\implies x=5$$

and so angle is $30^{\circ}$.

Answer 2

Note that $BE = CE$. If $H$ is the foot of the perp from $E$ to $BC$, $BH = BC/2$. Also, $\triangle BEH \sim \triangle CAD$.

If $BE = CE = x$,

$ \displaystyle \frac{CD}{AD} = \frac{BH}{BE} ~ \text { or } ~ \frac{3+x}{16} = \frac{5/2}{x}$

$x^2 + 3x - 40 = 0 \implies x = 5$

So, $\triangle BCE$ is equilateral and $\angle C = 60^\circ$. Hence $\angle A = 30^\circ$.