One Half is Green

In order to have the areas of $\Delta EAB$ and $\Delta ABC$ the same we will require that $|AE| = |AC|$ , since the two triangles share the same altitude to $EC.$ But since we require that $|AE| = |BD|$ we have that $|BD| = |AC|.$

Now the bisector length formula gives us that

$|BD|^{2} = \dfrac{4*3}{(4 + 3)^{2}}((4 + 3)^{2} - |AC|^{2})$ ,

which after substituting $|AC| = |BD|$ and simplifying yields that

$49*|AC|^{2} = 12(49 - |AC|^{2}) \Longrightarrow |AC|^{2} = \dfrac{588}{61}.$

Now by the Cosine rule we have that

$|AC|^{2} = |AB|^{2} + |BC|^{2} - 2|AB||BC|\cos(\angle B)$

$\Longrightarrow \dfrac{588}{61} = 9 + 16 - 24\cos(\angle B)$

$\Longrightarrow \cos(\angle B) = \dfrac{937}{1464} \Longrightarrow \angle B = \arccos(\dfrac{937}{1464}) = \boxed{50.206^{\circ}}$ to 3 decimal places.

This is just a minor application of cosine rule. And please specify weather answers is to be given in degrees or radians Guiseppi Butel