Q.Prove that a cyclic trapezium is an isosceles trapezium.

Given: ABCD is a cyclic trapezium where AD \(\parallel\) BC To prove: AB = DC Proof: \(\angle\)ADC + \(\angle\)DCB = 180° [∵ AD \(\parallel\) BC and DC is a transversal] Also, \(\angle\)BAD + \(\angle\)DCB = 180° [∵ ABCD is a cyclic quadrilateral] ∴ \(\angle\)ADC + \(\angle\)DCB = \(\angle\)BAD + \(\angle\)DCB ∴ \(\angle\)ADC = \(\angle\)BAD In triangles ∆BAD and ∆ADC: \(\angle\)BAD = \(\angle\)ADC \(\angle\)ABD = \(\angle\)DCA [Angles subtended by the same arc] AD is a common side ∆BAD ≅ ∆ADC [By A-A-S congruence rule] ∴ AB = DC (∵ Corresponding parts of congruent triangles) [Proved]
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