Christmas Streak 77/88: Geometry Rush #7

Oops, my hand slipped and I accidentally conveniently extended $\overline{\rm BD}$ and $\overline{\rm AC}.$

Since $\overline{\rm AD}$ bisects $\angle \rm BAC,$ we note that $\overline{\rm BD}=\overline{\rm DP},$ and thus $2\overline{\rm DE}=\overline{\rm CP}=10.$ ( $\rm \triangle BDE \sim \triangle BPC$ )

Then since $\overline{\rm AP}=\overline{\rm AB}=25,$ we figure out that $\overline{\rm AC}=\overline{\rm AP}-\overline{\rm CP}=15.$

$\triangle \rm DEF \sim \triangle ACF.$ So $\overline{\rm CF}=3\overline{\rm EF}=12.$

Then $\overline{\rm BC}=2\overline{\rm CE}=32.$

Therefore, the length of the perimeter of $\triangle \rm ABC$ is $\overline{\rm AB}+\overline{\rm BC}+\overline{\rm CA}=25+32+15=\boxed{72}.$

from bisector theorem and simliarity between $\triangle ACF$ and $\triangle DEF$ $\frac{AB}{BF} = \frac{25}{BF} = \frac{AC}{CF} = \frac{DE}{EF} = \frac{5}{4}$ $\therefore BF=20$

since $\triangle BDA$ is a triangle rectangle and $\angle ADP = \angle DAP$ thus $P$ is the midpoint and again by similiarity $\triangle BEP ~ \triangle BCA$ $2BE =BC = 32$

by triangle simliarity $\frac{FC}{FE} = \frac{12}{4} = \frac{AC}{ED} = \frac{AC}{5} = 3 \therefore AC = 15$ so answer is $15+ 25+ 32 = 72$