prove that a cyclic trapezium is always isosceles trapezium.

Prove that the line segment joining the mid-points of opposites sides of a quadrilateral bisect each other.

Please help me with these sums

**Prove that a cyclic trapezium is always isosceles trapezium.**

In a cyclic trapezium,

∠BAD + ∠BCD = 180°

∠BAD + ∠ABC = 180°

∴ ∠ABC = ∠BCD

In ∆ABC and ∆BDC,

⇒∠ABC = ∠BCD

⇒∠BAC = ∠CDB (angles in the same segment)

⇒BC = BC (common side)

⇒∆ABC ≅ ∆BDC

∴ AB = CD and AC = BD

Hence, cyclic trapezium ABCD is isosceles and diagonals are equal to each other.

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