Find $\angle E$

Since $AC=3$ , $AB=4$ , and $\angle A = 90^\circ$ , $BC=5$ . Let $\angle ACB = \alpha$ ; then $\tan \alpha = \dfrac 43$ , then

$\begin{aligned} \frac {2\tan \frac \alpha 2}{1-\tan^2 \frac \alpha 2} & = \frac 43 \\ 2\tan^2 \frac \alpha 2 + 3 \tan \frac \alpha 2 - 2 & = 0 \\ \left(2\tan \frac \alpha 2 - 1\right) \left(\tan \frac \alpha 2 + 2 \right) & = 0 & \small \blue{\text{SInce }\tan \alpha > 0} \\ \implies \tan \frac \alpha 2 & = \frac 12 \end{aligned}$

Drop a perpendicular from $E$ to the extension of $BC$ at $F$ . We note that $\triangle CEF$ and $\triangle CDA$ are similar. Then $\dfrac {CF}{AC} = \dfrac {CE}{CD} = 2$ , $\implies CF = 2 \times AC = 6$ . SInce $\tan \dfrac \alpha 2 = \dfrac {EF}{CF} = \dfrac 12$ , $\implies EF = 3$ . Then we have:

$\begin{aligned} \angle BEC & = \angle FEC - \angle FEB = \tan^{-1} \frac {CF}{EF} - \tan^{-1} \frac {BF}{EF} = \tan^{-1} \frac 63 - \tan^{-1} \frac 13 \\ \implies \tan \angle E & = \frac {2-\frac 13}{1+\frac 23} = 1 \\ \implies \angle E & = \boxed{45}^\circ \end{aligned}$

We have $|\overline {AC}|=3,|\overline {AB}|=4,|\overline {BC}|=\sqrt {3^2+4^2}=5,\sin \dfrac {C}{2}=\dfrac {1}{\sqrt 5}$

So, $|\overline {AD}|=\dfrac {3}{2},|\overline {BD}|=\dfrac {5}{2},|\overline {DC}|=\dfrac {3\sqrt 5}{2}$

Let $\angle {BEC}=α$ . Then

$\dfrac {5}{2\sin α}=\dfrac {\frac{3\sqrt 5}{2}}{\cos (\frac{C}{2}-α)}$

$\implies \tan α=1\implies α=\boxed {45\degree}$ .

By the Pythagorean Theorem, BC = 5. If angle(ACD) = r, we see tan(2*r) = 4/3. This means:

2 tan(r) / (1-tan(r) tan(r)) = 4/3, or

tan(r) = 1/2.

Now, using right triangle trigonometry, CD = 3*sqrt(5)/2 = ED and AD = 3/2. From this, we see BD = 4 - 3/2 = 5/2.

Now apply the Law of Cosines to triangle EBC to see:

(EB)^2 = 4 45/4 + 25 - 2 2 (3 sqrt(5)/2) 5 cos(r) = 10 (the last equality follows from cos(arctan(1/2)) = 2/sqrt(5).

Hence, EB = sqrt(10).

Now, note that angle(EDB) = 90 - r and apply the Law of Sines to triangle EDB to see:

sin(theta) = 1/sqrt(2)

Which means theta = 45.