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Question
ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.
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Solution
AB = AC ...[Given] ...(1)
AB = AD ...[Given] ...(2)
From (1) and (2), we have
AC = AD
Now, in ΔABC, we have
∠ABC + ∠ACB + ∠BAC = 180° ...[Angle sum property of a Δ]
⇒ 2∠ACB + ∠BAC = 180° ...(3) ...[∠ABC = ∠ACB (Angles opposite to equal sides of a Δ are equal)]
Similarly, in ΔACD,
∠ADC + ∠ACD + ∠CAD = 180°
⇒ 2∠ACD + ∠CAD = 180° ...(4) ...[∠ADC = ∠ACD (Angles opposite to equal sides of a Δ are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2∠ACD + ∠CAD = 180° +180°
⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒ 2∠BCD + 180° = 360° ...[∠BAC and ∠CAD form a linear pair]
⇒ 2∠BCD = 360° − 180°
⇒ 2∠BCD = 180°
⇒ ∠BCD = `(180°)/2`
Thus, ∠BCD = 90°
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