Log in to add a comment

Log in to add a comment

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.

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.