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Hint : circumcenter of an acute triangle
Hint2 : the incenter of triangle ABC is same as the circumcenter of triangle DEF
the vertices of the triangle DEF lie on the circle so lets assume that the triangle DEF is obtuse any obtuse triangle in a circle wuld be like-|dw:1443180458012:dw|
is this drawing appearing properly ?^
so now i have to prove that the center of the circle is inside DEF... but how?
well we know that the circle is tangent to triangle at D,E,F so the sides AB,BC,CA are tangents to the incircle the lines AB , BC, CA are represented by L1, L2,L3 but we can easily see that they do not form a triangle that keeps the circle in it ..
sorry... why not?
ok so lets extend the tangents and the point where they meet joing them would give us the triangle - |dw:1443181441216:dw|
What would I do if i wanted to calculate the angles of triangle DEF with the angles of triangle ABC?
yes, you can do that too!
You can do as follows: |dw:1443323977766:dw| By definition, the in-centre of a triangle is the intersection of the angle bisectors. Since the sides are tangent to the in-circle, the tangent lines from each vertex are equal. Hence \(\angle\)CDE =\(\angle\)CED= 90-\(\angle\)C÷2 Similarly \(\angle\)BFD =\(\angle\)BDF= 90-\(\angle\)B÷2 Consider angles at D on the side BC: \(\angle\)BDF+\(\angle\)FDE+\(\angle\)CED=180 (linear angles) Substitute angles BDF and CED: (90-C/2)+\(\angle\)FDE+(90-B/2)=180 simplify \(\angle\)FDE=(B+C)/2=(180-A)/2=90-A/2 => \(\angle\)FDE is acute since \(\angle\)A >0. The other two angles may be shown to be acute similarly.
oh i see ! Thanks!
Actually, it's even simpler than that! |dw:1443327095019:dw| Let O be the in-centre of the in-circle, then OF and OE are perpendicular to the tangents AF and AE. Thus \(\angle\)FOE+\(\angle\)FAE=360-180=180, or \(\angle\)FOE=180-\(\angle\)FAE But \(\angle\)FDE=(\(\angle\)FOE)/2=(180-\(\angle\)FAE)/2=90-(\(\angle\)A)/2 => \(\angle\)FDE is acute. The other two angles can be shown to be acute similarly.
that looks really neat !!
With the alternate segment theorem, it actually gets even more interesting. |dw:1443352742856:dw| Back to the problem: |dw:1443352833460:dw| AE and AF are tangent to the in-circle, so mAE=mAF. Hence for the isosceles triangle, \(\angle\)AFE=\(\angle\)AEF=90-A/2 By the alternate angle theorem, \(\angle\)FDE=\(\angle\)AEF=90-A/2 => \(\angle\)FDE is acute. The other angles can be shown similarly to be acute.