Theorem: A line parallel to one side of a triangle divides the other two proportionately. In the figure below, segment DE is parallel to segment BC and segment EF is parallel to AB:

The figure shows triangle ABC with segments DE and EF. Point D is on side AB, point E is on side AC, and point F is on side BC. Segment AD is 18, segment AE is 24, segment EC is 20, and segment FC is 30.

Which statement can be proved true using the given theorem?

Segment BF = 16
Segment BD = 20
Segment BD = 15
Segment BF = 32

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.

1 + (6 / x - 1)

step-by-step explanation:

divide using polynomial division.

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.

secant-secant formula:

(whole secant) x (external part) =           (whole secant) x (external part)

(8+y-2) *8=                                           (y+4+6) *6

(y+6)*8 = (y+10) *6

distribute

8y+48 = 6y +60

subtract 6y from each side

2y +48 = 60

subtract 48 from each side

2y = 12

divide by 2

y = 6

sv =     (y+4+6)

sv   = 6+4+6

sv = 16