Question

In triangle ABC shown, DE || BC.
a Class for Z00m
Test Player
x
Move options to the proof to show that DE divides AB and AC proportionally.
Statement
Reason
DE || BC
Given
LAS LA
Reflexive property
LABC * LADE
Corresponding angles of
are congruent.
A ABC - A ADE
Angle-angle similarity

Corresponding sides of similar triangles are
parallel lines
congruent triangles
AB AD
AC
=
AE
AB ≥ AC; AD = AE
congruent
proportional

Answer 1

To prove that line segment \( DE \) divides sides \( AB \) and \( AC \) proportionally in triangle \( ABC \) with \( DE \parallel BC \), we will use the properties of similar triangles as stated in your proof outline.

### Proof Structure

**Statement** | **Reason**
--- | ---
1. \( DE \parallel BC \) | Given
2. \( \angle ADE = \angle ABC \) and \( \angle ADE = \angle ACB \) | Corresponding angles formed by parallel lines \( DE \) and \( BC \)
3. \( \triangle ABC \sim \triangle ADE \) | Angle-Angle (AA) similarity criterion
4. \( \frac{AB}{AD} = \frac{AC}{AE} \) | Corresponding sides of similar triangles are proportional
5. \( AB \cdot AE = AC \cdot AD \) | Cross-multiplying the proportion from step 4

From steps 1 to 5, we've established that if \( DE \) is parallel to \( BC \), then the line segments \( AB \) and \( AC \) are divided proportionally. This means that:

\[
\frac{AB}{AD} = \frac{AC}{AE}
\]

This confirms that the segments \( AB \) and \( AC \) are indeed divided proportionally by segment \( DE \) in accordance with the properties of similar triangles.

### Summary
Thus, we have shown through the steps that \( DE \) divides \( AB \) and \( AC \) proportionally, confirming the statement required for your proof.