Question
✓
Answer
- 50º
Solution:
In the given $\angle\text{ABC},\ \text{AB}\ =\ \text{AC}$
Hence, $\angle\text{B} = \angle\text{C} = 40^\circ$
Now, by angle sum property,
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$\Rightarrow\ \angle\text{A} + 40^\circ + 40^\circ = 180^\circ$
$\Rightarrow\ \angle\text{A} = 100^\circ$
Now, in $\triangle\text{BAD}$ and $\triangle\text{DAC}.$
AB = AC (given)
BD = DC (D is the mid-point of BC)
$\angle\text{ABD} = \angle\text{ACD} = 40^\circ$
$\angle\text{A}= \frac{1}{2} × 100^\circ = 50^\circ $
Hence by SAS, $\triangle\text{BAD}$ and $\triangle\text{DAC}$ are congruent.
This means, $\angle\text{BAD} = \angle\text{CAD} = \frac{1}{2}$ Therefore, $\triangle\text{BAD}=50^\circ.$
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