Christmas Streak 71/88: Geometry Rush #1

Geometry Level 3

For a right triangle $\rm ABC$ with $\rm B = 90^{\circ},$ call $\rm D$ the foot of perpendicular from $\rm B$ to $\overline{\rm AC}.$

The intersections of the bisector of $\angle \rm A$ with $\overline{\rm BC}$ and $\overline{\rm BD}$ are $\rm E$ and $\rm F.$

Given that $\overline{\rm AB}=10$ and $\overline{\rm BF}=4,$ find the length of $\overline{\rm BE}.$

6 2 5 3 4

1 solution

Boi (보이)
Dec 16, 2017

Introducing a new point, $\rm H$ !

This revolutionary point will allow you to realize that $\angle \rm BEA = \angle \rm HEA$ along with $\overline{\rm EH}//\overline{\rm DF},$

so that $\rm \angle BEF = \angle AEH = \angle AFD = \angle BFE$ to show you blatantly that $\overline{\rm BE} = \overline{\rm BF} = \boxed{4}.$