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In a ΔABC, AB = AC and BA is produced to D such that AC = AD. Then the ∠BCD is :
Answer & Solution
Correct Answer:
Option
D
According to question,
ABC is an isosceles triangle.
∴ ∠C = ∠B = θ
∠CAD = ∠C + ∠B
∠CAD = θ + θ
∠CAD = 2θ
ADC is a isosceles triangle
∠C + ∠D + ∠A = 180°
2∠C = 180° - 2θ°
(∠C = ∠D)
∠C = 90° - θ
∴ ∠BCD = θ + 90 - θ
∠BCD = 90°
ABC is an isosceles triangle.
∴ ∠C = ∠B = θ
∠CAD = ∠C + ∠B
∠CAD = θ + θ
∠CAD = 2θ
ADC is a isosceles triangle
∠C + ∠D + ∠A = 180°
2∠C = 180° - 2θ°
(∠C = ∠D)
∠C = 90° - θ
∴ ∠BCD = θ + 90 - θ
∠BCD = 90°
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