Question : $\triangle ABC$ is an equilateral triangle. The side $BC$ is produced to point $D$. If $A$ joines $D$ and $B C=CD$, then the degree measure of angle $CAD$ is equal to:
Option 1: $30^{\circ}$
Option 2: $15^{\circ}$
Option 3: $45^{\circ}$
Option 4: $18^{\circ}$
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Correct Answer: $30^{\circ}$
Solution :
Given: BC = CD
And ABC is an equilateral triangle.
$\therefore$ AB = BC = CA
$\angle ACB = \angle ABC = \angle BAC = 60°$
To find $\angle CAD$
$\angle ACB + \angle ACD = 180°$
⇒ $ \angle ACD = 180°-60°$
⇒ $\angle ACD = 120°$
$\triangle ACD$ is an isosceles triangle
$\therefore$ $\angle CAD = \angle ADC$
$\angle CAD + \angle ACD + \angle ADC = 180°$
⇒ $\angle CAD + 120° + \angle ACD = 180°$
⇒ $2\angle CAD = 60°$
⇒ $\angle CAD = 30°$
Hence, the correct answer is $30^{\circ}$.
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