Prove that non-isosceles trapezium is not a cyclic

Hi Nandan,
Here is your answer,
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.
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.
 
Now,

• ∠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)
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,
∠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.
Hope it will clear your doubt about the topic

Regards