Can someone give me help me with this?
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I m gonna use ~ sign for angle
~CPD = y [Vertically opposite]
~ACB=~ADB=z/2 [Angle subtended by a chord on center is twice of that subtended by same chord on circumference]
~PCE=~PDE=180-(z/2) [Linear pair]
Sum of interior angle of quadrilateral CPDE will be 360
i.e. x + y + 180-(z/2) + 180-(z/2) = 360
x + y - z + 360 = 360 [z/2 + z/2 = z]
x+y=z
Hence proved.
Hope it helps :]
~CPD = y [Vertically opposite]
~ACB=~ADB=z/2 [Angle subtended by a chord on center is twice of that subtended by same chord on circumference]
~PCE=~PDE=180-(z/2) [Linear pair]
Sum of interior angle of quadrilateral CPDE will be 360
i.e. x + y + 180-(z/2) + 180-(z/2) = 360
x + y - z + 360 = 360 [z/2 + z/2 = z]
x+y=z
Hence proved.
Hope it helps :]
Chakshu290703:
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