Prove theorems using similarity

In the following triangle,

\frac{E C}{A E} = \frac{D B}{A D}

Below is the proof that

\overset{―}{E D} ∥ \overset{―}{C B}

. The proof is divided into two parts, where the title of each part indicates its main purpose.

Complete part B of the proof.

	Statement	Reason
1	$\frac{E C}{A E} = \frac{D B}{A D}$ ‍	Given
2	$1 + \frac{E C}{A E} = 1 + \frac{D B}{A D}$ ‍	Add $1$ ‍ to both sides. (1)
3	$\frac{A E}{A E} + \frac{E C}{A E} = \frac{A D}{A D} + \frac{D B}{A D}$ ‍	Rewrite $1$ ‍ as either $\frac{A E}{A E}$ ‍ or $\frac{A D}{A D}$ ‍. (2)
4	$\frac{A E + E C}{A E} = \frac{A D + D B}{A D}$ ‍	Add expressions with common denominator. (3)
5	$A C = A E + E C$ ‍	Segment addition postulate
6	$A B = A D + D B$ ‍	Segment addition postulate
7	$\frac{A C}{A E} = \frac{A B}{A D}$ ‍	Substitution (4,5,6)

	Statement	Reason
8	$∠ A ≅ ∠ A$ ‍	Reflexive property
9	$△ A E D \sim △$ ‍	similarity (Part A, 8)
10	$∠ 1 ≅ ∠ 2$ ‍	Measures of corresponding angles of similar triangles are equal. (9)
11	$\overset{―}{E D} ∥ \overset{―}{C B}$ ‍	If a transversal crosses two lines and angles are congruent, then the lines are parallel. (10)

High school geometry