In a ’ΔABC, AB = AC and BA is produced to D such that AC = AD. Then the ∠BCD is

∠ ABC = ∠ ACB = x ∠CAD = 180° - 2x ∴ ∠BAD = 180° - 2x ∴ 180° = (180° – 2x) × 2 ⇒ 180° – 2x = 90° ⇒ 2x = 90° = ∠ BCD Alternative: /

Let ∠ABC = 30°

$herefore angle mathrm{ABC}=angle mathrm{ACB}=30^{circ}$ (The opposite angles

corresponding to equal sides are equal)

$angle mathrm{BAC}=180^{circ}-(angle mathrm{ABC}+angle mathrm{ACB})$

$=180^{circ}-left(30^{circ}+30^{circ} ight)$

$=120^{circ}$

$angle mathrm{DAC}=180^{circ}-120^{circ}=60^{circ}$

$angle mathrm{ADC}=angle mathrm{ACD}(ecause mathrm{AC}=mathrm{AD})$

$herefore mathrm{In} riangle mathrm{ACD}$,

$angle mathrm{ACD}+angle mathrm{ADC}+angle mathrm{CAD}=180^{circ}$

$2 angle mathrm{ACD}=180^{circ}-60^{circ}$

$angle mathrm{ACD}=frac{120}{2}=60^{circ}$

$herefore angle mathrm{DCB}=angle mathrm{ACB}+angle mathrm{ACD}=30^{circ}+60^{circ}=90^{circ}$