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Minimum Angle Proof

Date: 07/05/2001 at 21:25:38
From: Pieta
Subject: Geometry involving incentres of triangles

Label the point of intersection of the angle bisectors of triangle ABC 
as Q. Let M be the midpoint of side BC. It is given that MQ = QA.
Find the minimum value of angle MQA.

I have tried to let the two lengths MQ and QA be the radii of a circle 
to see if angle MQA is subtended by any of the sides of the circle. I 
have also tried to use similar triangles and triangle congruences, but 
I do not seem to be making any progress. I have also tried to use the 
rule a/sinA = b/sinB, and I have tried to make use of the fact that 
the radius is absinC/( a+ b+ c).

Please help! :)

Date: 07/08/2001 at 08:44:19
From: Doctor Jaffee
Subject: Re: Geometry involving incentres of triangles

Hi Pieta,

This problem was challenging, and it took me a while to solve it, but 
I came up with a solution that works. Here is how I did it.

Draw triangle ABC and assume that side AB is shorter than side AC. If 
the opposite were true the proof would be similar.

Notice that triangle ABQ is congruent to triangle MBQ.  I'll leave 
that up to you to prove. The result of this congruency is that we now 
know that AB = BM, and since M is the midpoint of BC, we can conclude 
that AB = (1/2)AC.

                                sin C       sin A
According to the Law of Sines,  -----   =   -----
                                  AB         2AB

which means sin C = (1/2)sin A

Now, if you let the measure of angle A = 90 degrees, you can calculate 
the exact measure of angle AQM. If you let the measure of angle A be 
more or less than 90 degrees (call the measure 90 + x or 90 - x), then 
you can prove that the measure of angle AQM must be more than its 
measure when angle A is a right angle.

I have provided you with a bare outline of the proof. Try to fill in 
the details and if you want to check your solution with me, write 
back. If you are having difficulties, let me know and show me what you 
have done so far, and I'll try to help you some more.

Good luck.

- Doctor Jaffee, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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