Question

If ∠BAC = 17° and ∠CED = 17° are the two triangles, ΔBAC and ΔCED similar? If so, by what criterion?

Answer 1

No, not possible to tell that the the two triangles, ΔBAC and ΔCED

are similar ⇒ answer D

Step-by-step explanation:

Let us revise the cases of similarity

1. AAA similarity : two triangles are similar if all three angles in the first

  triangle equal the corresponding angle in the second triangle  

2. AA similarity : If two angles of one triangle are equal to the

   corresponding angles of the other triangle, then the two triangles  

   are similar.

3. SSS similarity : If the corresponding sides of the two triangles are

   proportional, then the two triangles are similar.

4. SAS similarity : In two triangles, if two sets of corresponding sides  

   are proportional and the included angles are equal then the two  

   triangles are similar.

In the two triangles BAC and CED

∵ m∠BAC = 17°

∵ m∠CED = 17°

∴ m∠BAC = m∠CED

But we need another pair of angles to prove that the two triangles are

similar by AA similarity criterion

OR

The lengths of sides BA , CA and CE , DE to show that

$\frac{BA}{CE}=\frac{CA}{DE}$ = constant ratio and prove that the

two triangles are similar by SAS similarity criterion

So it is not possible to prove that the two triangles are similar

No, not possible to tell that the the two triangles, ΔBAC and ΔCED

are similar

Learn more:

You can learn more about triangles in brainly.com/question/4354581

#LearnwithBrainly

Answer 2

Answer:

A) yes, by AA criterion

Step-by-step explanation:

If two angles in a triangle are congruent to two angles in another triangle then the two triangles are congruent. ∠BAC ≈ ∠CED since they have the same measure. ∠ACB ≈ ∠DCE since they are vertical angles. Since two angles are congruent, then you can conclude yes, by AA similarity criterion