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Question 8
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

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Solution


Consider a trapezium ABCD with AB | |CD and BC = AD.
Construction:
Draw AMCD and BNCD.

In ΔAMDandΔBNC
AD = BC (Given)
AMD=BNC (By construction, each angle is 90)
AM = BN (Perpendicular distance between two parallel lines is same)
ΔAMDΔBNC (RHS congruence rule)
ADC=BCD(CPCT)...(1)

BAD and ADC are on the same side of transversal AD.
BAD+ADC=180...(2)
BAD+BCD=180 [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.


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