Proving triangle similarities
Given line AD and line EC intersect at B

Prove triangle ABE is approximately triangle DBC

Hey there!

First of all we can look at the given statement and see that the first statement in the proof has a typo, it should say line AD and not line AC.

Anyways, let's continue:

1. Given. This is the given statement we have.

2. Vertical angle conjecture. Non-adjacent angles that are formed by two intersecting lines are always congruent, or equal in measure.

3. Proportion of corresponding sides. In the diagram, BC has a length of 5 and EB has a length of 15. This can be written as a ratio as a fraction, which can be reduced to 1/3.

4.   BD has a length of 4 and BA has a length of 12. Written as a fraction this can be reduced to 1/3.

5. Both are equal to 1/3, and 1/3 is equal to 1/3.

6. SAS/Side angle side theorem. If two corresponding sets of sides have the same proportions and the included angle is congruent, then the triangles are similar. In step 5 we prove the sides have the same proportions, and in step 2 we prove that the angle is congruent.

Thus proved.

Hope this helps!

15

step-by-step explanation:

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms