Question

What proportional segment lengths verify that BC¯¯¯¯¯∥DE¯¯¯¯¯ ? Fill in the boxes to correctly complete the proportion. $$ = $$ A triangle with vertices labeled as A, B, C. Side B C is base. Sides A B and A C contain midpoints D and E, respectively. A line segment is drawn from D to E. Side A D is labeled as 4. Side D B is labeled as 6. Side A E is labeled as 3.2. Side E C is labeled as 4.8.

Answer 1

answer is

4/6 then 3.2

took the test

Answer 2

Answer:

$\frac{4}{10}=\frac{3.2}{8.0}$

Step-by-step explanation:

Given,

ABC is a triangle,

In which, AD = 4 unit, DB = 6 unit, AE = 3.2 unit and EC = 4.8 unit,

⇒  $\frac{AD}{AB}=\frac{4}{4+6}=\frac{4}{10}=0.4$

$\frac{AE}{EC}=\frac{3.2}{3.2+4.8}=\frac{3.2}{8}=0.4$

$\implies \frac{AD}{AB}=\frac{AE}{EC}$

⇒ Δ ADE  ≅  Δ ABC  

⇒ ∠ADE ≅ ∠ABC  and ∠AED ≅ ∠ACB

By the converse of alternate interior angle theorem.

BC ║ DE

Hence, the required proportion of segment is,

$\frac{4}{10}=\frac{3.2}{8.0}$