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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

 Dec 14, 2018
 #1
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Here's an image :

 

 

 

Note that

 

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

 

 

cool cool cool

 Dec 14, 2018

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