CBSE Class 9 Answered
proove that
Asked by | 14 Mar, 2009, 04:32: PM
Expert Answer
Let that the angle bisectors of the said angles meet at P and ACD be the exterior angle of the base angle C.
So,
in triangle PBC,
angle PBC=1/2(B)...(as BP is the bisector of angle ABC)
angle PCB
=PCA+ACB
=1/2(angleACD)+C
=1/2(180-C)+C
=90-(C/2)+C
=90+(C/2)
So using angle sum property for triangle BPC, we get,
P+PBC+PCB=180
So,
P+1/2(B)+90+(C/2)=180
So,
P+[(B+C)/2]=90
So,
P+[(180-A)/2]=90.. (as B+C=180-A)
So,
P+90-(A/2)=90
So,
P=A/2
Hence proved.
Answered by | 14 Mar, 2009, 06:31: PM
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