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In triangle ABC, $\measuredangle$ ABC is bisected by $\overline{BE}$ and $\measuredangle$ ACB is bisected by $\overline{CD}$.
$\overline{BE} = \overline{CD} = x $. Show that $\overline{BD} = \frac{ac}{a + b}$ where $\overline{AB} = c$, $\overline{BC} = a$ and $\overline{AC} = b.$

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.

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Do you know of the angle bisector theorem? – Sawarnik Dec 11 '14 at 9:47

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

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