Question

1.prove that the angle between internal bisector of one base angle and the external bisectors of the other base angles of a triangle is equal to one-half of the vertical angle.
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Answer 1

Draw a triangle ABC. Let B and C are base angles and A be the vertical angle.  

Draw BX internal bisector of angle B.  

Draw CY external bisector of angle C.  

let BX and CY intersect at D.  

Now <BDC is the angle between internal bisector of one base angle (B) and external bisector of other base angle (C) of triangle ABC.  

from triangle DBC  

<BDC = 180 - [ <DBC + <DCB ]  

where <DBC is internally bisected angle of B and <DCB is <C + externally bisected angle of C  

<BDC = 180 - [ (B/2) + C + ( exterior angle of C /2 ) ]  

Recall that exterior angle of C = A + B  

so <BDC = 180 - [ (B/2) + C + (A + B) /2 ) ]  

=> <BDC = 180 - [ (B/2) + C + (A/2) + (B /2 ) ]  

=><BDC = 180 - [ B + C + (A/2) ]  

=> <BDC = [180 - ( B + C) ] - (A/2)  

=> <BDC = A - (A/2)  

=> <BDC = A/2