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.
Here's one method:
Find the center of the circle and call it "O"
Draw diameter F' D' through O such that F'D' is parallel to FD
Draw F'E and D'E .....
Now triangle F'ED' is a right triangle with angle F'ED' being the right angle [ since it intercepts the diameter ]
But the diameter in any circle is the greatest chord possible....so FD is shorter than F'D'
Therefore.....angle FED intercepts a lesser arc than angle F'ED'....therefore, angle FED < angle F'ED'
And since angle F'ED' was shown to be right, then angle FED must be < 90°....therefore, acute
And similarly, by drawing diameter F'E' parallel to FE and E'D' parallel to ED, we can show that angles FDE and EFD are also acute
I drew the diagram but it is kind of onfusing on how you labeled the angles. Can you provide a diagram?