Geometry

Quantitative Aptitude ( CCS)

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

Θ/2

2Θ

3Θ

Explanation:

In ΔABC,
Given, AB = BC
∠BCA = ∠BAC
And, AD = DC
∠CAD = ∠ACD
∴ ∠DAB = ∠DCB
∴ ∠DAB + ∠DCB = 180° ⇒ 2∠DAB = 180°
∴ ∠DAB = 90° = ∠DCB
In ΔADE, we know that
∴ ∠ADE = 180° – 90° – Θ = 90° – Θ
∴ ∠ADC = ∠ADE + ∠CDE = 2(90° – Θ)
∠ABC + ∠ADC = 180°
∠ABC = 180° – ∠ADC = 180° – 2(90° – Θ) = 2Θ

By: Amit Kumar

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