qr is produced to point S if d bisector of
Asked by | 17th Jan, 2010, 07:27: PM
in triangle PQR,
angle P+Q+R=180
So
P=180-(Q+R)....(i)
now
in triangle QRT,
angle TQR=Q/2
angle QRT=QRP+PRT
=R+[(1/2) (PRS)]
=R+[(1/2)(180-R)]
=(R/2)+90
In triangle QTR,
angle QTR+QRT+TQR=180
So,
QTR=180-{[(R/2)+90+(Q/2)]}
=90-[(R+Q)/2]
=90-[(180-P)/2].....from (i)
=P/2
Hence proved.
=
Answered by | 18th Jan, 2010, 08:21: AM
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