Geometry question about finding side length of a triangle

Question

I'm not getting anywhere with this question; I feel like I'm supposed to use similar triangles to find some side lengths but I haven't been able to find any. I let $EF=x$ and found that $AP=PD=\frac{5}{8}x$, but haven't gotten any further than that. Can someone please help me out?

Answer 1

You're nearly there! Similar triangles / side length ratios is all that you need.

Hint: $BPD, BEF$ are also similar, so $BD:DF = ?:?$.

Hint: $ABD, CAD$ are also similar.

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$BD:DF:FC = 5:3:12$, which means that $BD : DC = 1:3$
$ ABD, CAD$ are similar $\Rightarrow BD : AD : DC = 1 : \sqrt{3} : 3$.
So $ \angle C = 30^ \circ$ and $AD = 15 \sin 30^ \circ$.

Answer 2

Hint: use barycentric method. In any triangle:

In our triangle $BC$ is divided by $D$ as $1:3$, therefore $C$ is $30$ degrees

For method details see the article I wrote several years ago.