Answer :

Given: ΔABC, external angle bisectors of ACB, line parallel to BC passing through point A meets


To prove:


Let CD be the external angle bisector of ACB, i.e., CD is angle bisector of ACE. And let line l be the parallel line to side BC through point A. So line l and CD meets at point D. The figure of the above question is as shown below,



Now in ΔABC,


We know in a triangle the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.


So in this case,


ACE = ABC + BAC………(i)


But it is also given CD is angle bisector of ACE, so


ACD = DCE


ACE = 2ACD


Substituting the above value in equation (i), we get


2ACD = ABC + BAC……..(ii)


Now in ΔABC,


We know in a triangle the sum of all three interior angles is equal to 180°.


So in this case,


BAC + ABC + ACB = 180°


Substituting the value from equation (ii) in above equation, we get


2ACD + ACB = 180°


2ACD = 180° - ACB




Hence proved


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