Question

Gary is using an indirect method to prove that segment DE is not parallel to segment BC in the triangle ABC shown below:

Answer 1

Answer:  4:10 ≠ 6:14

Step-by-step explanation:

Given : ABC is a triangle,

In which $D\in AB$ and $E\in AC$

We have to prove that : DE is not parallel to BC,

That is, DE ∦ BC

Proof: Let us assume that,

DE is parallel to segment BC.

Thus, By the corresponding angle theorem,

∠ ADE ≅ ∠ ABC and ∠ AEC ≅ ∠ ACB

By AA similarity theorem,

$\triangle ADE\sim \triangle ABC$

Thus, By the property of similar triangles,

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

Given AD = 4, DB=6 ⇒ AB = AD+DB = 4+6 = 10

AE = 6, EC = 8 ⇒ AC = AE + EC = 6+8 = 14

But, $\frac{4}{10} \neq \frac{6}{14}$

Thus, triangle ADE is not similar to ABC,

That is, our assumption is wrong.

⇒ DE is not parallel to BC

Therefore, First Option is correct.

Answer 2

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Based on the situation above the inequality will he use to contradict the assumption is 
4:10 ≠ 6:14
if DE is parallel to BC
then
4: (4+5) = 6 : (6 + 8)