In $\triangle ABC$, bisectors of $\angle B$ and $\angle C$ meet opposite sides at $D$ and $E$. If $\angle B<\angle C$, show $CE<BD$.

Question

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

Using the fact that $\angle B<\angle C$, 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 $CF<BF$, we only have to prove that $FE<DF$. 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?

Answer 1

Use theorem of sines on triangles $\triangle ABD$ and $\triangle ACE$ to obtain: $$\dfrac{BD}{\sin A} = \dfrac{AB}{\sin\left( A+\frac B2\right)},\quad \dfrac{CE}{\sin A} = \dfrac{AC}{\sin\left(A+\frac C2\right)}.$$ Combining these two yields: $$\dfrac{BD}{CE} = \dfrac{AB}{AC}\cdot\dfrac{\sin\left( A+\frac C2\right)}{\sin\left( A+\frac B2\right)}$$ and the rest should be easy.