# Prove that non-isosceles trapezium is not a cyclic

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 )

Now,

- ∠ADF = ∠BCE ( By CPCT i.e. Corresponding Parts of Congruent Triangles )

∠ADC = ∠BCD (equation 1)

Also,

- ∠CBE = ∠DAF ( By CPCT )

∠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

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