Question

the following two - column proof proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally.
statement:

1. $de\parallel ac$
2. $ab$ is a transversal that intersects two parallel lines.
3. $\angle bde\cong\angle bac$
4. $\angle b\cong\angle b$

5.

1. $\frac{bd}{ba}=\frac{be}{bc}$

reason:

1. given
2. conclusion from statement 1
3. corresponding angles postulate
4. reflexive property of equality

5.

1. converse of the side - side - side similarity theorem

which statement and reason accurately completes the proof?
options:

1. $\triangle bde\sim\triangle bac$; angle - angle (aa) similarity postulate
2. $\triangle bde\sim\triangle bac$; angle - angle (aa) similarity postulate
3. $\triangle bde\sim\triangle bac$; side - angle - side (sas) similarity postulate
4. $\triangle bde\sim\triangle bac$; side - angle - side (sas) similarity postulate

Explanation:

1. First, we know that $DE\parallel AC$ (given). When a transversal ($AB$) intersects two parallel lines ($DE$ and $AC$), the corresponding angles are equal. So $\angle BDE\cong\angle BAC$ (corresponding angles postulate) and $\angle B\cong\angle B$ (reflexive property of congruence, which is the same angle for both triangles).
2. For triangle similarity, the Angle - Angle (AA) Similarity Postulate states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. Here, we have two pairs of congruent angles ($\angle BDE\cong\angle BAC$ and $\angle B\cong\angle B$) between $\triangle BDE$ and $\triangle BAC$, so by AA similarity, $\triangle BDE\sim\triangle BAC$. The other options use SAS similarity, but we don't have information about the sides being proportional in the way required for SAS (we are using angle - angle here), so the AA similarity postulate with the statement $\triangle BDE\sim\triangle BAC$ is the correct one.

Answer:

1. $\triangle BDE \sim \triangle BAC$; Angle - Angle (AA) Similarity Postulate