given that △abc is equilateral, and $\overline{ad}$ bisects $\angle a$, show that the side lengths of △adb are in a ratio of $1 : \sqrt{3} : 2$.
$\checkmark$ 2. by definition of an angle bisector, $m\angle bad = m\angle cad = 30^\circ$
$\checkmark$ 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. $bd = cd$
6. $2(\quad) = bc$
options for 6: ab, bd, ac

Question

Accepted Answer

# Explanation:
## Step1: Recall properties of equilateral triangle
In an equilateral triangle $ \triangle ABC $, all angles are $ 60^\circ $, and all sides are equal ($ AB = BC = AC $). Since $ \overline{AD} $ bisects $ \angle A $, by the angle - bisector definition, $ \angle BAD=\angle CAD = \frac{60^\circ}{2}=30^\circ $. Also, in an equilateral triangle (which is a special case of an isosceles triangle), the angle bisector of the vertex angle is also the perpendicular bisector of the base. So, $ AD\perp BC $, which means $ \angle BDA = 90^\circ $, and $ BD = CD $ (because $ AD $ bisects $ BC $).

## Step2: Let the side length of $ \triangle ABC $ be $ 2x $
Let $ AB = BC=AC = 2x $. Since $ BD = CD $ and $ BC = 2x $, then $ BD=x $.

## Step3: Apply the Pythagorean theorem in $ \triangle ADB $
In right - triangle $ \triangle ADB $, we know that $ AB $ is the hypotenuse, $ BD $ is one leg, and $ AD $ is the other leg. By the Pythagorean theorem $ AB^{2}=AD^{2}+BD^{2} $. Substituting $ AB = 2x $ and $ BD=x $ into the formula:
$$
\begin{align*}
(2x)^{2}&=AD^{2}+x^{2}\
4x^{2}&=AD^{2}+x^{2}\
AD^{2}&=4x^{2}-x^{2}\
AD^{2}&=3x^{2}\
AD&=\sqrt{3}x
\end{align*}
$$

## Step4: Find the ratio of the sides of $ \triangle ADB $
The sides of $ \triangle ADB $ are $ BD=x $, $ AD = \sqrt{3}x $, and $ AB = 2x $. To find the ratio, we divide each side by $ x $ (since $ x
eq0 $). So the ratio of $ BD:AD:AB=x:\sqrt{3}x:2x = 1:\sqrt{3}:2 $.

For question 6: Since $ BD = CD $ and $ BC=BD + CD $, we have $ BC=2BD $, so $ 2(BD)=BC $.

# Answer:
For the ratio part: The side lengths of $ \triangle ADB $ are in the ratio $ 1:\sqrt{3}:2 $ as shown above. For question 6, the answer is $ BD $, so $ 2(BD)=BC $.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall properties of equilateral triangle

Step2: Let the side length of \( \triangle ABC \) be \( 2x \)

Step3: Apply the Pythagorean theorem in \( \triangle ADB \)

Step4: Find the ratio of the sides of \( \triangle ADB \)

Answer: