in the figure ABCD is a trapezium . if the vertices are on a circle, proove that it is an iscoseles trapezium​

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1. 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 that helps 🙂

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