Let ABC be a triangle, $\measuredangle$ ABC is bisected by $BE$ and $\measuredangle$ ACB is bisected by $CD$.
It's given that $BE = CD$. Show that $BD = \frac{AB \cdot BC}{AC + BC}$.
A high school student asked me this question today and I was completely stumped. He goes to a magnet school in the city. The topics he went over in class were Stewart's Theorem and Ptolemy's Theorem.
For convenience let $a = BC, b = AC , c = AB$
Since $$\dfrac{BD}{DA}=\frac{\triangle CBD}{\triangle CDA}=\frac{CB\cdot CD\sin\angle BCD}{CD\cdot CA\sin\angle DCA}=\frac{BC}{CA}=\frac ab$$ therefore $$BD=c\cdot\frac a{a+b}$$ Q.E.D