In a quadrilateral ABCD bisectors of angle A and angle B meet at O. Prove that angle AOB =1/2(
Asked by masara
| 4th Dec, 2009,
09:58: PM
Expert Answer:
Sum of the angles of a quadrilateral is 360
so A+
B+
C+
D = 360
C+
D = 360 -
A+
B = 2(180 - (
A+
B )/2) ---(i)
Also in triagle AOB
AOB+
ABO+
BAO = 180
AOB + 1/2(
A+
B ) =180
AOB = 180 - 1/2(
A+
B ) -------(ii)
from (i) and (ii)
we have AOB =1/2 (
C+
D )
Answered by
| 5th Dec, 2009,
10:26: AM
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