Question : In an isosceles triangle $ABC$, $AB = AC$ and $\angle A = 80^\circ$. The bisector of $\angle B$ and $\angle C$ meet at $D$. The $\angle BDC$ is equal to:
Option 1: $90^\circ$
Option 2: $100^\circ$
Option 3: $130^\circ$
Option 4: $80^\circ$
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Correct Answer: $130^\circ$
Solution :
Given: $ABC$ is an isosceles triangle and $AB = AC$.
Let $\angle B=\angle C=x$
$\angle A+\angle B+\angle C = 180^\circ$
⇒ $80^\circ+x+x = 180^\circ$
⇒ $2x = 180^\circ-80^\circ$
⇒ $2x = 100^\circ$
⇒ $x = 50^\circ$
So, $\angle B = \angle C = 50^\circ$
The bisector of $\angle B$ and $\angle C$ meet at $D$.
So, we have $\angle CBD = \angle BCD=\frac{50^\circ}{2} = 25^\circ$
In $\triangle BDC$,
$\angle BDC + \angle CBD + \angle CDB = 180^\circ$
⇒ $\angle BDC + 25^\circ + 25^\circ = 180^\circ$
⇒ $\angle BDC = 180^\circ - 50^\circ$
⇒ $\angle BDC = 130^\circ$
Hence, the correct answer is $130^\circ$.
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