Question

If non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer 1

According to the question,

We have,

ABCD is a trapezium in which AD||BC

Non-parallel sides AB and DC of the trapezium ABCD are equal i.e.,

AB = DC.

To prove: Trapezium ABCD is cyclic.

Construction: Draw AM and DN such that they are perpendicular on BC.

Proof: In right triangles AMB and DNC,

∠AMB = ∠DNC = 90^o

AB = DC [Given]

Since perpendicular distance between two parallel lines are same,

AM = DN

ΔAMB ≅ ΔDNC [By RHS congruence rule]

∠B = ∠C [CPCT]

And ∠1 = ∠2

∠BAD = ∠1+ 90

= ∠2 + 90

= ∠CDA

Now, in quadrilateral ABCD

∠B +∠C +∠CDA+∠BAD = 360

∠B +∠B +∠CDA +∠CDA = 360

2(∠B +∠CDA) = 360

∠B +∠CDA =180

We know that,

If any pair of opposite angles of a quadrilateral is 180^o, then the quadrilateral is cyclic.

Hence, the trapezium ABCD is cyclic.