proove that

Asked by  | 14th 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  | 14th Mar, 2009, 06:31: PM

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