Mathematics
Answer
Let ABCD be the cyclic trapezium in which AB || DC, AC and BD are the diagonals.
It's seen that chord AD subtends ∠ABD and chord BC subtends ∠BDC at the circumference of the circle.
But, ∠ABD = ∠BDC [Alternate angles are equal]
∴ Chord AD must be equal to chord BC [As equal chord subtends equal angles at circumference]
⇒ AD = BC.
Hence, proved that in a cyclic trapezium, the non-parallel sides are equal.
Related Questions
Prove that the angle subtended at the centre of a circle is bisected by the radius passing through the mid-point of the arc.
In the given figure, P is the mid-point of arc APB and M is the mid-point of chord AB of a circle with centre O. Prove that:
(i) PM ⟂ AB
(ii) PM produced will pass through the centre O
(iii) PM produced will bisect the major arc AB.
If P is a point on a circle with centre O. If P is equidistant from the two radii OA and OB, prove that arc AP = arc PB.
In the given figure, two chords AC and BD of a circle intersect at E. If arc AB = CD, prove that : BE = EC and AE = ED.
