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To prove that any given quadrilateral is cyclic, we need to prove that its opposite angles are supplementary (i.e. they add up to 180˚).
take an isosceles trapezium:

(Here AB and CD are parallel and AD = BC )
We need to prove that ∠BAD + ∠BCD = 180 and ∠ADC + ∠ABC = 180˚.

Start by constructing two perpendiculars, AF and BE, from segment AB to segment CD.

Now, in ΔADF and ΔBCE,
∠AFD = ∠BEC (by construction - both are right angles at the feet of the perpendiculars)
AD = BC (property of trapezium)
AF = BE ( perpendicular distance between two parallel lines remains same irrespective of the point of measurement along the line )
Thus, ΔADF ≅ ΔBCE by RHS ( Right angle - Hypotenuse - Side ) congruency.
∠ADF = ∠BCE ( By CPCT i.e. Corresponding Parts of Congruent Triangles )Since ∠ADF is same as ∠ADC and ∠BCE is same as ∠BCD,
∠ADC = ∠BCD (equation 1)
∠CBE = ∠DAF ( By CPCT )Adding the right angles ∠ABE and ∠BAF to the above angles,
∠CBE + ∠BAF = ∠CBE + ∠ABE
Thus, ∠ABC = ∠BAD (equation 2)

So, adding equations 1 and 2,
∠ADC + ∠ABC = ∠BCD + ∠BAD
Since the sum of all the angles in a quadrilateral is 360˚,
∠ADC + ∠ABC + ∠BCD + ∠BAD = 360˚
2 (∠ADC + ∠ABC) = 2 (∠BCD + ∠BAD) = 360˚
∠ADC + ∠ABC =∠BCD + ∠BAD = 180˚

Since the opposite angles are supplementary, an isosceles trapezium is a cyclic quadrilateral.
Hence proved.
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