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Question
In isosceles triangle ABC, AB = AC. The side BA is produced to D such that BA = AD.
Prove that: ∠BCD = 90°
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Solution
Const: Join CD.
In ΔABC,
AB = AC ...[ Given ]
∴ ∠C = ∠B .......(i) [angles opp. to equal sides are equal]
In ΔACD,
AC = AD ...[Given]
∴ ∠ADC = ∠ACD ...(ii)
Adding (i) and (ii)
∠B + ∠ADC = ∠C + ACD
∠B + ∠ADC = ∠BCD ...(iii)
In ΔBCD,
∠B + ∠ADC + ∠BCD = 180°
∠BCD + ∠BCD = 180° ...[From (iii)]
2∠BCD = 180°
∠BCD = 90°
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