Let the incircle of triangle ABC be tangent to sides BC, AC, and AB at D, E, and F, respectively. Prove that triangle DEF is acute.
+981 Assuming you already have the diagram, here is what I got:
Since angle EDF is an inscribed angle, we have:
\(\angle{EDF}=\frac{{\text{arc}\ EF}}{2} \)
Since angle A is formed by two tangents, we find:
\(\angle{A}=(\text{arc}\ EDF - \text{arc}\ EF)/2=(360º-2\cdot\text{arc}\ EF)/2\)
Therefore, \(\angle{A}=180º-\text{arc}\ EF = 180º - 2 \angle{EDF}\)
Solving for EDF, we get \(\angle{EDF}=90º-{\angle{A}}/2\)
Since we know angle A is greater than zero, angle EDF is less than 90. Similarly, other angles are also acute.
I hope this helped,
Gavin
