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.

Answer 1

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°