Consider A, B, C as the points
Taking B as the centre and radius equal to AC construct an arc
Taking C as the centre and radius equal to AB construct another arc which cuts the arc at D
So we get D as the required point BD and CD
Consider △ ABC and △ DCB
We know that
AB = DC and AC = DB
BC and CB are common i.e. BC = CB
By SSS congruence criterion
△ ABC ≅ △ DCB
∠BAC = ∠CDB (c. p. c. t)
We know that BC subtends equal angles ∠BAC and ∠CDB on the same side
So we get A, B, C, D are cyclic.
Therefore, the points A, B, C, D are cyclic.