# Incircle and Tangency Proof

Let the incircle of triangle $ABC$ be tangent to sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ at $D$, $E$, and $F$, respectively. Prove that triangle $DEF$ is acute.

I can't prove that the triangles are similar, and we are not given any specific angle of side measurements. I'm stuck, any answer is greatly appreciated.

• Hint: calculate angle $\widehat{EDF}$ for example. – dxiv Sep 22 '16 at 0:25

$CDE$ is an isosceles triangle, hence $\widehat{EDC}=\frac{\pi-C}{2}$. In a similar way, $\widehat{DBF}=\frac{\pi-B}{2}$, hence: $$\widehat{FDE}=\pi-\frac{\pi-B}{2}-\frac{\pi-C}{2}=\frac{B+C}{2}$$ and since $B+C<\pi$, $\widehat{FDE}$ is an acute angle. The same applies to $\widehat{FED}$ and $\widehat{DFE}$.
• What do you mean by $B+C<\pi$? – Yuna Kun Sep 22 '16 at 0:57
• @YunaKun: that the amplitude of the $B$ angle, plus the amplitude of the $C$ angle, is less than $180^\circ$ because $ABC$ is a triangle. – Jack D'Aurizio Sep 22 '16 at 0:59
• You may compute the squared lengths of the sides of $DEF$ and apply the converse Pythagorean theorem, if you are happy with that, instead of using that the sum of the angles of a triangle is $180^\circ$. – Jack D'Aurizio Sep 22 '16 at 1:08