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anonymous

  • one year ago

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  1. ganeshie8
    • one year ago
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    |dw:1443180472134:dw|

  2. ganeshie8
    • one year ago
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    Hint : circumcenter of an acute triangle

  3. ganeshie8
    • one year ago
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    Hint2 : the incenter of triangle ABC is same as the circumcenter of triangle DEF

  4. ytrewqmiswi
    • one year ago
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    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|

  5. ytrewqmiswi
    • one year ago
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    is this drawing appearing properly ?^

  6. anonymous
    • one year ago
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    so now i have to prove that the center of the circle is inside DEF... but how?

  7. ytrewqmiswi
    • one year ago
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    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 ..

  8. anonymous
    • one year ago
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    sorry... why not?

  9. ytrewqmiswi
    • one year ago
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    ok so lets extend the tangents and the point where they meet joing them would give us the triangle - |dw:1443181441216:dw|

  10. Loser66
    • one year ago
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    |dw:1443181774267:dw|

  11. anonymous
    • one year ago
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    What would I do if i wanted to calculate the angles of triangle DEF with the angles of triangle ABC?

  12. mathmate
    • one year ago
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    yes, you can do that too!

  13. mathmate
    • one year ago
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    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.

  14. anonymous
    • one year ago
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    oh i see ! Thanks!

  15. mathmate
    • one year ago
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    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.

  16. ganeshie8
    • one year ago
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    that looks really neat !!

  17. mathmate
    • one year ago
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    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.

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