ABC is a triangle with no obtuse angles. The perpendicular AD from A to BC meets the circumcircle of the triangle at E. If the incentre of the triangle is equidistant from A and E, prove that .

I have no idea how to begin.

thanks

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- February 27th 2010, 07:39 AMarzeTriangle proof
ABC is a triangle with no obtuse angles. The perpendicular AD from A to BC meets the circumcircle of the triangle at E. If the incentre of the triangle is equidistant from A and E, prove that .

I have no idea how to begin.

thanks - February 28th 2010, 12:13 AMsa-ri-ga-ma
If I is the incenter, and AI = IE, then I must lie on the perpendicular bisector of AE and it must pass through circumcenter of ABC.

Secondly I must be equidistant from all the three sides of the triangle ABC.

These two conditions are satisfied only when ABC is**equilateral**triangle. In that case B = C = 60 degrees and cosB + cosC = 1. - February 28th 2010, 11:26 AMGrandad
Hello arzeThis is

*much, much*harder than the other trig questions you've posted. I wonder where it came from.

I'm afraid thisisn't true. The triangle ABC needn't be equilateral.

I have a proof - see below - but it's very complicated for such an apparently simple result. Perhaps someone may be able to come up with something more straightforward?

Here goes.

Look at the attached diagram. I is the incentre.

I haven't drawn the circumcircle - the only property of this that I'm using is that:(angles in same segment) Call this equation (1)Then we need to express in two ways:

1. In terms of the sides and angles of any triangle .So first:

2. Using the fact that is isosceles, with .

...(2)Then, expressing using the fact that is isosceles:

...(3)

, from (1), from (3)

, from (2)

, using . Call this equation (4)

The radius of the incircle, , is given by:, where is the area of the triangle ABCNow bisects . So

...(5)

, from (5)

You would think that there was an easier way, but I can't see it!

Grandad - February 28th 2010, 01:26 PMOpalg
I also found this a tough problem. I have a cumbersome solution that I held back from posting in the hope that someone would find a neater way of doing it. No offence to Grandad, but his solution is about as cumbersome as mine. Just for the record, here's an outline of how I did it.

The incentre I is on the perpendicular bisector of AE. So is the circumcentre O. So the line IO is parallel to BC (see the attached modified version of Grandad's diagram). My proof is based on the fact that the triangles BOC and BIC have the same base BC and the same vertical height, and therefore the same area.

The area of BOC is , where R is the circumradius and*a*is the length BC. But . Also, and . So the area of triangle BOC is .

The area of BIC is . By the sine rule, . So the area of triangle BIC is .

Equate those two areas to see that . On the left side, the numerator is and the denominator is . Make those substitutions and cross-multiply (noticing that cancels in the two denominators):

Now use the fact that to write this as , which simplifies to .

Phew! (Whew) - February 28th 2010, 03:51 PMarze
Thanks everyone! The proofs were really long.

Grandad, I got this from an A-level pure math textbook.