QQuestion
Prove that a cyclic trapezium is an isosceles trapezium.
✓Answer
ABCD is a cyclic trapezium of which AD || BC
AB = DC or ABCD is a rectangle and AC = BD.
∠ADC + ∠DCB = 180° [∠ AD || BC and DC is transversal]
Again, ∠BAD + ∠DCB = 180° [∠ ABCD is a cyclic quadrilateral]
∴ ∠ADC + ∠DCB = ∠BAD + ∠DCB ∴ ∠ADC = ∠BAD ..
In ∆BAD and ∆ADC, ∠BAD = ∠ADC [From (1) we get]
∠ABD = ∠DCA [∴ Angles in the same segment]
AD is common side
∴ ∆BAD = ∆ADC [A-A-S congruence property]
∴ AB = DC .. ABCD is an isosceles trapezium or a rectangle and AC = BD (Similar part of congruent triangle) [Proved]
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