+781 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?
+569 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
+781 I don't quite understand why Cphill chose to rearrange the equation like that though.
+1490 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.
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 
( This is one way to prove it...This may not be the answer they're looking for.)
+781 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.
