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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.

Guest May 17, 2018
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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

 
GYanggg  May 17, 2018

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