Angle bisectors AX and BY of triangle ABC intersect at point I. Prove that angle XIB=90 degrees-angle bca/2.
Please help. I drew another angle bisector of angle BCA and named it Z
Here's an image :
BAC + ABC + BCA = 180 ..... so ......
BAC/2 + ABC/2 + BCA/2 = 90 ⇒ [90 - BCA/2 = BAC/2 + ABC/2] (1)
And by the Exterior Angle Theorem.....
XIB = BAC/2 + ABC/2 (2)
Sub (1) into (2) and we have that
XIB = 90 - BCA/2