Prove that non-isosceles trapezium is not a cyclic

Hi Nandan,
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˚).

Here’s 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. • ∠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.

Now,
• ∠ADF = ∠BCE ( By CPCT i.e. Corresponding Parts of Congruent Triangles )
Since ∠ADF is same as ∠ADC and ∠BCE is same as ∠BCD,
Also,
• ∠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,
Since the sum of all the angles in a quadrilateral is 360˚, 