Question

given that $\triangle abc$ is equilateral, and $\overline{ad}$ bisects $\angle a$, show that the side lengths of $\triangle adb$ are in a ratio of $1 : \sqrt{3} : 2$.
\checkmark 3. equilateral triangles are isosceles triangles.
\checkmark 4. because the bisector of the vertex angle of an isosceles triangle is the $\perp$ bisector of the base, $m\angle bda = m\angle cda = 90^\circ$.
\checkmark 5. $bd = cd$
\checkmark 6. $2(bd) = bc$

1. triangles $adb$ and $adc$ are $30^\circ - 60^\circ - 90^\circ$ triangles.

let $bd = 1$. then $bc = $ dropdown with 1, 2, 3, which also means that $ab = 2$.
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image of triangle $abc$ with $ad$ as altitude to $bc$ at $d$

Explanation:

Step1: Recall BD and BC relationship

From step 6, we know that \( 2(BD) = BC \).

Step2: Substitute BD = 1

Given \( BD = 1 \), substitute into the formula: \( BC = 2\times1 = 2 \).

Answer:

2