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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!

 Jun 13, 2019
 #1
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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

 

 

cool cool cool

 Jun 13, 2019
 #2
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Thank you so much, CPhill!!!!!!!!!!!

Guest Jun 14, 2019

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