given that △abc is equilateral, and $\overline{ad}$ bisects $\angle a$, show that the side lengths of △adb are in a ratio of $1 : \sqrt{3} : 2$.
1. equilateral triangles are also equiangular, so $m\angle a = m\angle b = m\angle c = 60^\circ$.
2. by definition of an angle bisector, $m\angle bad = m\angle cad = 30^\circ$
3. equilateral triangles are isosceles triangles.
4. because the bisector of the vertex angle of an isosceles triangle is the $\perp$ bisector of the base, $m\angle bda = m\angle cda = 90^\circ$.
5. bd = $
\begin{matrix}
ab \\
ac \\
cd
\end{matrix}
$ check

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Accepted Answer

# Explanation:
## Step1: Recall properties of equilateral triangle
In an equilateral triangle $ \triangle ABC $, $ AB = BC = AC $, and $ AD $ is the angle bisector (also median and altitude as per properties of isosceles/equilateral triangle). So $ D $ bisects $ BC $, meaning $ BD = CD $.
## Step2: Determine the length of $ BD $
Since $ D $ is the mid - point of $ BC $, $ BD=\frac{1}{2}BC $. But in equilateral triangle $ AB = BC = AC $, and from the property of the angle bisector (which is also the median) in an equilateral triangle, $ BD = CD $. So among the options $ AB $, $ AC $, $ CD $, we know that $ BD = CD $ because $ D $ divides $ BC $ into two equal parts.

# Answer:
CD

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QUESTION IMAGE

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Explanation:

Step1: Recall properties of equilateral triangle

Step2: Determine the length of \( BD \)

Answer: