jmc

First of all, let us make a sketch, although it is far from necessary: A $13:14:15$ triangle is a Heronian triangle, or a triangle that has integer sides and integer area. This can easily be verified using Heron's Formula. In fact, it is easy to find that a $13:14:15$ triangle is merely two $9:12:15$ and $5:12:13$ right triangles mashed together at the common leg.

Regardless, the first step is finding the area of the triangle. Since the perimeter is $13+14+15=42,$ we have that $s=21.$ Therefore, $$\begin{array}{rl}[△ABC]& =\sqrt{s(s-a)(s-b)(s-c)}\\ & =\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21\cdot 8\cdot 7\cdot 6}\\ & =\sqrt{7\cdot 3\cdot 4\cdot 2\cdot 7\cdot 3\cdot 2}=\sqrt{7^2\cdot 3^2\cdot 2^4}\\ & =7\cdot 3\cdot 2^2=84.\end{array}$$By the Angle Bisector Theorem, we know that $BD:DC=AB:AC=13:15.$ That means that the area of $△ABD$ to $△ADC$ must also have the $13:15$ ratio, and that means the ratio of $[△ADC]:[△ABC]$ is $15:28.$

Then, $[△ADC]=\frac{15}{28}\cdot [△ABC]=\frac{15}{28}\cdot 84=\boxed{45}.$