Question

triangle bac was dilated from triangle bde at a scale factor of 2. what proportion proves that \\(\tan\angle d = \tan\angle a\\)? (1 point)

Explanation:

Step1: Identify dilated side lengths

Since the scale factor is 2:
$BC = 2 \times BE = 2 \times 3 = 6$
$BA = 2 \times BD = 2 \times 2 = 4$

Step2: Define tangent for each angle

For $\tan\angle D$ (in $\triangle BDE$):
$\tan\angle D = \frac{BE}{BD} = \frac{3}{2}$
For $\tan\angle A$ (in $\triangle BAC$):
$\tan\angle A = \frac{BC}{BA} = \frac{6}{4}$

Step3: Simplify and set proportion

Simplify $\frac{6}{4} = \frac{3}{2}$, so the proportion is:
$\frac{BE}{BD} = \frac{BC}{BA}$

Answer:

$\boldsymbol{\frac{3}{2} = \frac{6}{4}}$ (or simplified as $\boldsymbol{\frac{BE}{BD} = \frac{BC}{BA}}$)