Question 132

A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC ⊥ BD, ∠CAD = θ. Then the angle ∠ABC =

Solution

AB = BC and AD = DC , $$\angle CAD = \theta$$

In isosceles $$\triangle$$ADC

=> $$\angle$$CAD + $$\angle$$ACD + $$\angle$$CDA = 180

=> $$2\theta$$ + $$\angle$$CDA = 180

=> $$\angle$$CDA = 180 - $$2\theta$$

Sum of opposite angles in a cyclic quadrilateral = 180

=> $$\angle$$CDA + $$\angle$$ABC = 180

=> $$\angle$$ABC + 180 - $$2\theta$$ = 180

=> $$\angle$$ABC = $$2\theta$$

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