Question

(03.06 mc)
in \\( \triangle a b c \\) shown below, \\( \frac{b d}{b a}=\frac{b e}{b c} \\)

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:

which reason can be used to fill in the numbered blank space?
\\( \bigcirc 1. \triangle a b c \sim \triangle b e d \\)

1. side-angle-side similarity postulate

\\( \bigcirc 1. \triangle a b c \sim \triangle b e d \\)

1. side-side-side similarity theorem

\\( \bigcirc 1. \triangle a b c \sim \triangle d b e \\)

1. side-angle-side similarity postulate

\\( \bigcirc 1. \triangle a b c \sim \triangle d b e \\)

1. side-side-side similarity theorem

Explanation:

Step1: Analyze the triangles

We have \(\frac{BD}{BA}=\frac{BE}{BC}\) (given) and \(\angle B=\angle B\) (reflexive property).

Step2: Apply similarity criterion

By the Side - Angle - Side (SAS) similarity postulate, if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the triangles are similar. Here, for \(\triangle ABC\) and \(\triangle DBE\), \(\frac{BD}{BA}=\frac{BE}{BC}\) and \(\angle B\) is common. So, \(\triangle ABC\sim\triangle DBE\) by the Side - Angle - Side Similarity Postulate.

Answer:

1. \(\triangle ABC\sim\triangle DBE\)
2. Side - Angle - Side Similarity Postulate (So the correct option is: 1. \(\triangle ABC\sim\triangle DBE\), 2. Side - Angle - Side Similarity Postulate)