Rebecca Villa

2022-07-09

In $\mathrm{△}ABC$, $\mathrm{\angle }B<\mathrm{\angle }C$, and $BD$ and $CE$ are angle bisectors. $D$ is on $AC$ and $E$ is on $AB$. Prove that $CE.

Using the fact that $\mathrm{\angle }B<\mathrm{\angle }C$, I got that $AB>AC$ and $BF>CF$ (where $F$ is the intersection of the two angle bisectors).
I then expressed $CE=CF+FE$ and $BD=BF+DF$. Since $CF, we only have to prove that $FE. I tried to use the angle bisector theorem to get some relations that might help me prove that, but here's where I got stuck. Could someone please help me out?

Jenna Farmer

Expert

Use theorem of sines on triangles $\mathrm{△}ABD$ and $\mathrm{△}ACE$ to obtain:
$\frac{BD}{\mathrm{sin}A}=\frac{AB}{\mathrm{sin}\left(A+\frac{B}{2}\right)},\phantom{\rule{1em}{0ex}}\frac{CE}{\mathrm{sin}A}=\frac{AC}{\mathrm{sin}\left(A+\frac{C}{2}\right)}.$
Combining these two yields:
$\frac{BD}{CE}=\frac{AB}{AC}\cdot \frac{\mathrm{sin}\left(A+\frac{C}{2}\right)}{\mathrm{sin}\left(A+\frac{B}{2}\right)}$
and the rest should be easy.

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