Question : $ABC$ is an isosceles triangle with $AB = AC$, The side $BA$ is produced to $D$ such that $AB = AD$. If $\angle ABC = 30^{\circ}$, then $\angle BCD$ is equal to:
Option 1: $45^{\circ}$
Option 2: $90^{\circ}$
Option 3: $30^{\circ}$
Option 4: $60^{\circ}$
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Correct Answer: $90^{\circ}$
Solution :
Given that $ABC$ is an isosceles triangle, $AB = AC$, $AB = AD$ and $\angle ABC = 30^{\circ}$.
Such that,
In $\triangle ABC,$
$\angle ACB =\angle ABC$ (Angles opposite to equal sides of a triangle are also equal)
In $\triangle ACD,$
$\angle ADC =\angle ACD$ (Angles opposite to equal sides of a triangle are also equal)
In $\triangle BCD,$
$⇒\angle BCD=\angle ACB+\angle ACD$
Now, $\angle ABC+\angle BCD+\angle ADC=180^{\circ}$ (Angle sum property)
$⇒\angle ACB+\angle ACB+\angle ACD+ \angle ACD=180^\circ$
$⇒2(\angle ACB+\angle ACD)=180^\circ$
$⇒2\angle BCD=180^{\circ}$
$\therefore\angle BCD=90^{\circ}$
Hence, the correct answer is $90^{\circ}$.
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