Find $k$

$AB = 12$

$AD = 15$

$EC = k$

Project $E$ orthogonally onto $AC$ , its image being point $F$ .

$\angle BAE=\angle FAE=\theta$

$\triangle ABE \cong \triangle AFE$

$AF = AB = 12$

$FD = AD - AF = 15 - 12 = 3$

Since $EF$ is the altitude on the hypotenuse of right triangle $AED$ , $EF$ is the geometric mean of $AF$ and $FD$ .

$EF^2 = (AF)(FD)$

$EF^2 = (12)(3)$

$EF = 6$

$BE = EF = 6$

$\tan \theta = \dfrac{BE}{AB} =\dfrac{1}{2}$

$\tan 2\theta=\dfrac{2\tan \theta}{1-\tan^2 \theta}=\dfrac{2 \times \dfrac{1}{2}}{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{4}{3}$

$\tan 2\theta=\dfrac{6+k}{12}=\dfrac{4}{3}$

$\dfrac{6+k}{12}=\dfrac{4}{3}$

$48=18+3k$

$\boxed{k=10}$

Let $\angle BAC = \angle CAD =\theta$ . Then $AC = \dfrac {12}{\cos \theta}$ and:

$\begin{aligned} \frac {AC}{\cos \theta} & = AD \\ \frac {12}{\cos^2 \theta} & = 15 \\ \implies \cos^2 \theta & = \frac {12}{15} = \frac 45 \\ \cos \theta & = \frac 2{\sqrt 5} \\ \cos (2\theta) & = 2\times \frac 45 - 1 = \frac 35 \end{aligned}$

Therefore, $k = BE-BC = 12 \tan 2\theta - 12 \tan \theta = 12 \times \dfrac 43 - 12 \times \dfrac 12 = \boxed{10}$ .

Let the two equal angles be $α$ each, length of the bisector line be $y$ , and length of the base of the triangle whose one segment is $k$ be $x+k$ . Then

$\cos α=\dfrac{12}{y}=\dfrac{y}{15}\implies y^2=180$ . So, $180=144+x^2\implies x=6$ .

Hence, $\tan α=\dfrac{6}{12}=0.5, \tan 2α=\dfrac{6+k}{12}=\dfrac{2\tan α}{1-\tan^2 α}=\dfrac{4}{3}\implies k=16-6=\boxed {10}$ .