Calculate Angle BAD in an Isosceles Triangle with Median AD

Question

ABC is an isosceles triangle.

AB = AC

AD is the median.

Calculate the size of angle $∢\text{BAD}$

00:07 Let's find the angle B A D.

00:10 This triangle is isosceles, as the information states.

00:15 We know A D is a median from the given details.

00:19 In an isosceles triangle, the median also splits the angle exactly in half.

00:25 So, angle B A D is half of angle B A C.

00:32 And that's how we solve it!

To determine $\angle \text{BAD}$ in triangle $ABC$ , follow these steps:

The problem states that $ABC$ is an isosceles triangle where $AB = AC$ .
$AD$ is given as the median, which also acts as the angle bisector because $ABC$ is isosceles.
Since $\angle BAC = 70^\circ$ , and $AD$ bisects $\angle BAC$ , $\angle BAD = \angle CAD = \frac{\angle BAC}{2}$ .
Therefore, $\angle BAD = \frac{70^\circ}{2} = 35^\circ$ .

Thus, the measure of $\angle \text{BAD}$ is $\mathbf{35^\circ}$ .