It is given that ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD
Construct DE ⊥ AB and CF ⊥ AB
Consider △ ADE and △ BCF
We know that
∠AED + ∠BFC = 90o
From the figure it can be written as
∠ADE = ∠ADC – 90o = ∠BCD – 90o = ∠BCF
It is given that
AD = BC
By AAS congruence criterion
△ ADE ≅ △ BCF
∠A = ∠B (c. p. c. t)
We know that the sum of all the angles of a quadrilateral is 360o
∠A + ∠B + ∠C + ∠D = 360o
By substituting the values
2 ∠B + 2 ∠D = 360o
By taking 2 as common
2 (∠B + ∠D) = 360o
By division
∠B + ∠D = 360/2
So we get
∠B + ∠D = 180o
So, ABCD is a cyclic quadrilateral.
Therefore, it is proved that the points A, B, C and D lie on a circle.