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In triangle ABC, angle bisectors AD and BE meet at point I. Prove that angle DIB = 90 - angle BCA/2.

 

 

What I've got currently is the diagram, and that 

Since triangle ABC is a triangle, its angles add up to 180, so angleABC+angle BCA+CAB=180. When we divide this equation by two we get

\(\angle{ABC}+\angle{BCA}+\angle{CAB}=180^\circ\\ \frac{\angle{ABC}}{2}+\frac{\angle{BCA}}{2}+\frac{\angle{CAB}}{2}=90^\circ\).

 

Then I don't know how to proceed from here...any guiding points or help?

 Jul 26, 2020
 #1
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To be honest with you, geometry is not my game, so I will not try. (I am willing to, but it may be wrong anyways.) Instead of having a sloppy answer by me, check out this answer by CPhill, since this problem has been posted before. 

 

CPhill answered very well, and it was quite easy to follow along : https://web2.0calc.com/questions/help-quick-plz_2 

 Jul 26, 2020
 #2
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I don't quite understand why Cphill chose to rearrange the equation like that though. 

gwenspooner85  Jul 26, 2020
 #4
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I mean, as you can see he did almost the same approach as you, gwenspooner85, so I think you should be able to decipher his thinking

iamhappy  Jul 26, 2020
 #3
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In triangle ABC, angle bisectors AD and BE meet at a point I. Prove that angle DIB = 90 - angle BCA/2.

 

I'm purposely using a right-angled triangle because it gives me a better overall picture.smiley 

 

Angle ADB = 180º - (90º + 30º) = 60º

 

Angle  DIB = 90º ∠BCA /2               ∠DIB = 90º - 30º/2 = 75º

 

Look at a triangle BID                       ∠DIB = 180º - (45º + 60º) = 75º  wink

 

( This is one way to prove it...This may not be the answer they're looking for.)laugh

 

 Jul 26, 2020
edited by Dragan  Jul 26, 2020
 #5
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Ah...I see, thank you. I think I prefer this way. But instead of using numbers, using variables would provide an overall proof for any angle degrees values.

gwenspooner85  Jul 26, 2020

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