Angle bisectors \(\overline{AX}\) and \(\overline{BY}\) of \(\triangle ABC\) intersect at point ** L. **Prove that \(\angle XLB = 90^\circ - \frac{\angle BCA}{2}\).

Thank you and please help!

Guest Jun 13, 2019

#1**+3 **

Look at the following diagram :

Let CD = the bisector of angle BCA

We want to prove that XLB = 90 - BCA/2

And BCA/2 = (1/2)C = XCL

So

XLB = 90 - XCL

XLB + XCL = 90

Prove that XLB + XCL = 90

A+ B + C = 180

(1/2)A + (1/2)B + (1/2)C = 90

XLB = YLA (vertical angles)

YLA + ALB= 180

BAL + ABL + ALB = 180

Therefore

YLA + ALB = BAL + ABL + ALB (subtract ALB from both sides)

YLA = BAL + ABL

YLA = (1/2)A + (1/2)B

So....by substitution

[(1/2)A + (1/2)B] + (1/2)C = 90

[ YLA ] + (1/2)C = 90

[ XLB ] + (1/2)C = 90

[ XLB ] + [ XCL] = 90

CPhill Jun 13, 2019