Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 75

Let the incircle of triangle ABC be tangent to sides

and at D, E, and F, respectively. Prove that triangle DEF is acute.I just have no idea how to do this. I think I should find the angles in terms of the larger triangles angles, but I duno how...:/

this thing is haaard. thanks!!!:D

Good. You can read.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,466

hi thedarktiger,

Like the avatar by the way.

Diagram below.

Some preliminaries.

The centre O is on the intersection of the angle bisectors.

In triangles AFO and AEO, AO is common, OF = OE = radius and FAO = EAO so these are congruent. Therefore, AF = AE.

Similarly, BF = BD and CE = CD.

So triangles AEF, BFD and CDE are each isosceles.

Now to get an expression for angle EFD.

AFE = 90 - A/2 and BFD = 90 - B/2

=> EFD = 180 - (90 - A/2 + 90 - B/2) = (A + B)/2 = (180 - C)/2 = 90 - C/2

As C < 180 => C/2 < 90 => EFD is acute.

A similar argument can be used for the other two angles of DEF.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 75

thanks!

Good. You can read.

Offline

Pages: **1**