In the figure below, ABCD is a cyclic quadrilateral, AB = BC and angle BAC = 70, then angle ADC is

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Q: 148 (IAS/2001)
In the figure below, ABCD is a cyclic quadrilateral, AB = BC and angle BAC = 70°, then angle ADC is

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,5,6,1,3,2,5

keywords: 

{'cyclic quadrilateral': [0, 0, 1, 0], 'angle bac': [0, 0, 1, 0], 'angle adc': [0, 0, 1, 0], 'abcd': [0, 0, 1, 4], 'ab': [0, 0, 1, 0], 'figure': [0, 1, 1, 0], 'bc': [1, 0, 1, 2]}

In a cyclic quadrilateral, the sum of the opposite angles is 180 degrees. If ∠BAC is 70 degrees and AB is equal to BC, triangle ABC is isosceles and thus ∠ABC is also 70 degrees. The sum of the two opposite angles of quadrilateral ABCD is ∠ABC + ∠ADC = 180 degrees. By substituting the known angle ⦟ABC = 70 degrees, we have ∠ADC = 180 - 70 = 110 degrees.

Option 1 suggests that ∠ADC is 40 degrees, but this would imply that the sum of the opposite angles is less than 180 degrees, which is not possible in a cyclic quadrilateral.

Option 2 says that ∠ADC is 80 degrees, however, this is not possible as our calculation shows that ∠ADC should be 110 degrees.

Option 3 suggests ∠ADC = 110 degrees. This is incorrect, as it incorrectly assumes that ∠ABC (70 degrees) and ∠ADC equal each other, but this goes against the idea that the two opposite angles in a cyclic quadrilateral should sum to 180 degrees.

Option 4 presents that ∠ADC = 140 degrees. The mistake

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