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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.

 May 17, 2018

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,



 May 17, 2018

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