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 have tried proving that triangle DEF's angles were les than the opposite angles in triangle ABC, but that wasn't really complete. Can anyone help?
What do all the dollar signs ($) signify?
They're just a format you can ignore them
haha nice try maight
To prove that triangle $DEF$ is acute, we need to show that all three angles of triangle $DEF$ are less than $90^\circ$. One way to approach this proof is by utilizing contradiction.
Assume, for the sake of contradiction, that triangle $DEF$ is not acute, which means at least one angle is obtuse or right. Since $D$ is the point of tangency on side $BC$ of triangle $ABC$, we can consider the points $D$, $B$, and $C$ in this order and apply the Inscribed Angle Theorem.
Considering the arc $DB$ (which is intercepted by $DE$) and the arc $DC$ (which is intercepted by $DF$), we can observe the following:
1. If either $\angle DFE$ or $\angle DEB$ is obtuse or right, then the intercepted arc $DB$ or $DC$ would be obtuse or right. However, by the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc, which implies that either $\angle ABC$ or $\angle ACB$ would be obtuse or right.
2. If both $\angle DFE$ and $\angle DEB$ are obtuse or right, then the intercepted arc $DB$ would be greater than a semicircle. However, this would imply that $\angle ABC + \angle ACB > 180^\circ$, which is not possible in a triangle.
Therefore, by contradiction, we can conclude that triangle $DEF$ must be acute (i.e., all three angles of triangle $DEF$ are acute).