#### Question

In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.

#### Solution

A cyclic trapezium ABCD in which AB ∥ DC and AC and BD are joined.

To prove-

(i) AD = BC

(ii) AC = BD

Proof :

∵ chord AD subtends ∠ABD and chord BC subtends BDC

At the circumference of the circle.

But ∠ABD = ∠BDC [proved]

Chord AD = Chord BC

⇒ AD = BC

Now in ∆ADC and ∆ BCD

DC = DC [Common]

∠CAD = ∠CBD [angles in the same segment]

And AD = BC [proved]

By Side – Angle – Side criterion of congruence, we have

∴ ∆ADC ≅ ∆BCD [ SAS axion]

The corresponding parts of the congruent triangle are congruent

∴ AC = BD [c.p.c.t]

Is there an error in this question or solution?

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In a Cyclic-trapezium, the Non-parallel Sides Are Equal and the Diagonals Are Also Equal. Prove It. Concept: Arc and Chord Properties - Angles in the Same Segment of a Circle Are Equal (Without Proof).

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