Question

In a cyclic-trapezium, the non-parallel sides are equal and the
diagonals are equal. Prove it.
[Circle]​

Answer 1

Step-by-step explanation:

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 ΔADC and Δ BCD DC = DC [Common] ∠CAD = ∠CBD [angles in the same segment] And AD = BC [proved] By Side – Angle – Side criterion of congruence, we have ΔADC ≅ ΔBCD [ SAS axion] The corresponding parts of the congruent triangle are congruent Therefore,  AC = BD [c.p.c.t]Read more on Sarthaks.com - https://www.sarthaks.com/157463/in-cyclic-trapezium-the-non-parallel-sides-are-equal-and-the-diagonals-are-also-equal-prove