prove that the angle contained by the bisectors of any two consecutive angles of a quadrilateral is equal to half the sum of the remaining angles
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tp: AOB=(x+y)/2
In Quad. ABCD, 2z+2a+x+y= 360°
2(z+a)= 360-(x+y)
z+a= 180- (x+y)/2........(1)
In triangle AOB, AOB=180-(z+a)
=180-180+(x+y)/2........(from 1)
so AOB= (x+y)/2
In Quad. ABCD, 2z+2a+x+y= 360°
2(z+a)= 360-(x+y)
z+a= 180- (x+y)/2........(1)
In triangle AOB, AOB=180-(z+a)
=180-180+(x+y)/2........(from 1)
so AOB= (x+y)/2
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