Question

If in $△ABCD$ is any point on $BC$ such that $\frac{AB}{AC}=\frac{BD}{DC}$ and $\angle B={70}^o$, $\angle C={50}^o$, then find $\angle BAD$

Answer 1

In $△ABC$
$\angle B={70}^o$, $\angle C={50}^o$
$\angle A+\angle B+\angle C={180}^o$
$\Rightarrow$ $\angle A={180}^o-(\angle B+\angle C)$
$\Rightarrow$ $\angle A={180}^o-({70}^o+{50}^o)$
$\Rightarrow$ $\angle A={180}^o-{120}^o$
$\Rightarrow$ $\angle A={60}^o$
Thus, $\frac{AB}{AC}=\frac{BD}{DC}$ (given)
$△ABD\sim △ADC$
$\Rightarrow$ $\angle BAD$ and $\angle DAC$ will be equal to each other
$2\angle BAD={60}^o$
$\angle BAD=\frac{60}{2}$
$\angle BAD={30}^o$