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!
+128460 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
