Question
In fig. $\triangle\text{ABC}$ is a triangle such that $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}},\angle\text{B}=70^\circ,\angle\text{C}=50^\circ.$ Find the $\angle\text{BAD}.$
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Answer
In $\triangle\text{ABC},$$\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}},\angle\text{B}=70^\circ,\angle\text{C}=50^\circ$
We know that,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+70^\circ+50^\circ=180^\circ$
$\Rightarrow\angle\text{A}+120^\circ=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-120^\circ$
$\Rightarrow\angle\text{A}=60^\circ$
$\because\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\therefore$ AD is bisector of $\angle\text{A}$ so,
$\angle1=\angle2$
$\Rightarrow\angle\text{A}=\angle1+\angle2$
$\Rightarrow60^\circ=2\angle1$
$\Rightarrow\angle1=\frac{60^\circ}{2}$
$\Rightarrow\angle1=30^\circ$
$\Rightarrow\angle\text{BAD}=30^\circ$
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