In ΔABC shown below, BD over BA equals BE over BC:

Triangle ABC with segment DE intersecting sides AB and BC respectively.

The following flowchart proof with missing statements and reasons proves that if a line intersects two sides of a triangle and divides these sides proportionally, the line is parallel to the third side:

Top path, by Given the ratio of line segments BD to BA is equal to the ratio of line segments BE to BC. By Side Angle Side Similarity Postulate, triangle ABC is similar to triangle DBE. By Corresponding Parts of Similar Triangles, angle BDE is congruent to angle BAC. By Converse of the Corresponding Angles Postulate, line segment DE is parallel to line segment AC. Bottom path, by space labeled as 2, space labeled by 1 occurs. By Side Angle Side Similarity Postulate, triangle ABC is similar to triangle DBE. By Corresponding Parts of Similar Triangles, angle BDE is congruent to angle BAC. By Converse of the Corresponding Angles Postulate, line segment DE is parallel to line segment AC.

Which reason can be used to fill in the numbered blank space?

1. ∠A ≅ ∠A

2. Reflexive Property of Equality

1. ∠A ≅ ∠B

2. Corresponding Parts of Similar Triangles

1. ∠A ≅ ∠B

2. Corresponding Angles Postulate

1. ∠B ≅ ∠B

2. Reflexive Property of Equality

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .