anonymous
  • anonymous
***
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ganeshie8
  • ganeshie8
|dw:1443180472134:dw|
ganeshie8
  • ganeshie8
Hint : circumcenter of an acute triangle
ganeshie8
  • ganeshie8
Hint2 : the incenter of triangle ABC is same as the circumcenter of triangle DEF

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ytrewqmiswi
  • ytrewqmiswi
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|
ytrewqmiswi
  • ytrewqmiswi
is this drawing appearing properly ?^
anonymous
  • anonymous
so now i have to prove that the center of the circle is inside DEF... but how?
ytrewqmiswi
  • ytrewqmiswi
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 ..
anonymous
  • anonymous
sorry... why not?
ytrewqmiswi
  • ytrewqmiswi
ok so lets extend the tangents and the point where they meet joing them would give us the triangle - |dw:1443181441216:dw|
Loser66
  • Loser66
|dw:1443181774267:dw|
anonymous
  • anonymous
What would I do if i wanted to calculate the angles of triangle DEF with the angles of triangle ABC?
mathmate
  • mathmate
yes, you can do that too!
mathmate
  • mathmate
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.
anonymous
  • anonymous
oh i see ! Thanks!
mathmate
  • mathmate
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.
ganeshie8
  • ganeshie8
that looks really neat !!
mathmate
  • mathmate
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.