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anonymous
 one year ago
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anonymous
 one year ago
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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443180472134:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Hint : circumcenter of an acute triangle

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Hint2 : the incenter of triangle ABC is same as the circumcenter of triangle DEF

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.2the 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 likedw:1443180458012:dw

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.2is this drawing appearing properly ?^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now i have to prove that the center of the circle is inside DEF... but how?

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.2well 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 ..

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.2ok so lets extend the tangents and the point where they meet joing them would give us the triangle  dw:1443181441216:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What would I do if i wanted to calculate the angles of triangle DEF with the angles of triangle ABC?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1yes, you can do that too!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1You can do as follows: dw:1443323977766:dw By definition, the incentre of a triangle is the intersection of the angle bisectors. Since the sides are tangent to the incircle, 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: (90C/2)+\(\angle\)FDE+(90B/2)=180 simplify \(\angle\)FDE=(B+C)/2=(180A)/2=90A/2 => \(\angle\)FDE is acute since \(\angle\)A >0. The other two angles may be shown to be acute similarly.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Actually, it's even simpler than that! dw:1443327095019:dw Let O be the incentre of the incircle, then OF and OE are perpendicular to the tangents AF and AE. Thus \(\angle\)FOE+\(\angle\)FAE=360180=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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that looks really neat !!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1With 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 incircle, so mAE=mAF. Hence for the isosceles triangle, \(\angle\)AFE=\(\angle\)AEF=90A/2 By the alternate angle theorem, \(\angle\)FDE=\(\angle\)AEF=90A/2 => \(\angle\)FDE is acute. The other angles can be shown similarly to be acute.
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