In the given figure, ABCD is a quadrilateral in which AD = BC, AB||CD and ∠ADC = ∠BCD. Show that the points A,B,C and D lie on a circle.

Given ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD.

To prove ABCD is a cyclic quadrilateral.

Construction Draw DE perpendicular to AB and CF perpendicular to AB.

Proof In ∆AED and ∆BFC,

∠ AED = ∠BFC=90° [∵ DE ⊥ AB and CE ⊥ AB]

ED = FC

[ ∵ distance between two parallel lines are equal ]

and AD =BC [given]

Then, ∠A = ∠B [by CPCT] … ,(i)

and ∠ADE = ∠BCF [by CPCT]

Since, ABCD is a quadrilateral.

∠A+∠B+∠C+∠D=360°

[∵ sum of all angles of a quadrilateral is 360°]

⇒∠B + ∠B + ∠D + ∠D = 360° [from Eq. (i)]

⇒2 ∠B + 2 ∠D=360°

⇒∠B + ∠D=180°

Hence, ABCD is a cyclic quadrilateral because the sum of opposite angles is 180°.