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.