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A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC ⊥ BD, ∠CAD = θ. Then the angle ∠ABC = (SSC CGL 1st Sit. 2013)
- a)θ
- b)2/θ
- c)2θ
- d)3θ
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC ⊥ B...
∠B + ∠D = 180°
∠A + ∠C = 180°
∠DAC = θ {given}
∠AED = 90° {given}
In ΔAED,
∴ ∠ADE = 90° – θ = ∠CDE
∴ ∠ABC = 180° – 2 (90° – θ) = 2θ
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Community Answer
A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC ⊥ B...
In a cyclic quadrilateral, the opposite angles are supplementary, so ∠ABC + ∠ADC = 180°.
Since AB = BC, we can conclude that ∠ABC = ∠BCA.
Similarly, since AD = DC, we can conclude that ∠ADC = ∠ACD.
Therefore, we have ∠ABC + ∠ADC = ∠BCA + ∠ACD = 180°.
Since ∠BCA + ∠ACD = 180°, we can conclude that ∠BCA = ∠ACD = 90°.
Thus, we have proven that the diagonals of the cyclic quadrilateral are perpendicular to each other.
Since AB = BC, we can conclude that ∠ABC = ∠BCA.
Similarly, since AD = DC, we can conclude that ∠ADC = ∠ACD.
Therefore, we have ∠ABC + ∠ADC = ∠BCA + ∠ACD = 180°.
Since ∠BCA + ∠ACD = 180°, we can conclude that ∠BCA = ∠ACD = 90°.
Thus, we have proven that the diagonals of the cyclic quadrilateral are perpendicular to each other.
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A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC ⊥ BD, ∠CAD = θ. Then the angle ∠ABC = (SSC CGL 1st Sit. 2013)a)θb)2/θc)2θd)3θCorrect answer is option 'C'. Can you explain this answer?
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