In figure below, $Δ\ ABC$ is a triangle such that $\frac{AB}{AC}\ =\ \frac{BD}{DC}$, $∠B\ =\ 70^o$, $∠C\ =\ 50^o$, find $∠BAD$.
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Given:

In the given figure, $Δ\ ABC$ is a triangle such that $\frac{AB}{AC}\ =\ \frac{BD}{DC}$, $∠B\ =\ 70^o$ and $∠C\ =\ 50^o$.

To do:

We have to find the measure of $∠BAD$.

Solution:

In $\vartriangle ABC$,

$\angle A+\angle B+\angle C=180^o $  [Angle sum property of a triangle]

$\angle A=180^o-(70+50)^o$

$\angle A=60^o$

$\frac{AB}{AC} = \frac{BD}{DC}$  (given)

Therefore, AD is the angle bisector of $\angle A$.

This implies,

$\angle BAD=\frac{\angle A}{2} =\left(\frac{60}{2}\right)^o =30^o $

The measure of $\angle BAD$ is $30^o $.

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Updated on: 10-Oct-2022

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