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

A triangle ABC is shown. D is a point on side AB and E is a point on side AC. Points D and E are joined using a straight line. The length of AD is equal to 4, the length of DB is equal to 6, the length of AE is equal to 6 and the length of EC is equal to 8.

He starts with the assumption that segment DE is parallel to segment BC.

Which inequality will he use to contradict the assumption?

4:10 ≠ 6:14
4:6 ≠ 6:14
4:10 ≠ 6:8
4:14 ≠ 6:10

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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)

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parmesanchilliwack parmesanchilliwack · Answer 2 · 2019-01-07T12:37:32+01:00

Answer: 4:10 ≠ 6:14

Step-by-step explanation:

Given : ABC is a triangle,

In which [tex]D\in AB[/tex] and [tex]E\in AC[/tex]

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,

[tex]\triangle ADE\sim \triangle ABC[/tex]

Thus, By the property of similar triangles,

[tex]\frac{AD}{AB} = \frac{AE}{AC}[/tex]

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

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

But, [tex]\frac{4}{10} \neq \frac{6}{14}[/tex]

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.