Question

In triangle ABC , side AB is 6 and side AC is 4: Which statement is needed to prove that segment DE is parallel to segment BC and half its length? a. Segment AD is 3, and segment AE is 4. b. Segment AD is 3, and segment AE is 2. c. Segment AD is 12, and segment AE is 8. d. Segment AD is 12, and segment AE is 4.

Answer 1

Answer:

Segment AD is 3, and segment AE is 2.

Step-by-step explanation:

In a triangle, the line joining the mid points of two sides is parallel and half of the third sides of the triangle.

Here, ABC is a triangle,

In which,

AB = 6,

AC = 4,

D∈ AB and E∈AC

Let DE ║BC,

And, $DE=\frac{1}{2}BC$

In triangles ADE and ABC,

$\angle ADE\cong \angle ABC$  ( Alternative interior angle theorem )

$\angle AED\cong \angle ACB$

By AA similarity postulate,

$\triangle ADE\sim \triangle ABC$

∵ Corresponding sides of similar triangle are in same proportion,

$\implies \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$

$\frac{AD}{AB}=\frac{AE}{AC}=\frac{BC}{2BC}$

$\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{2}$

$\implies AD = \frac{AB}{2}\text{ and }AE =\frac{AC}{2}$

$\implies AD = \frac{6}{2}=3\text{ and }AE =\frac{4}{2}=2$

Hence, the correct option would be,

Segment AD is 3, and segment AE is 2.