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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