Question

prove the side - angle - side similarity theorem.
given: \\(\frac{ab}{qr}=\frac{ac}{qs}\\), \\(\angle a\cong\angle q\\)
prove: \\(\triangle abc\sim\triangle qrs\\)

statements

1. choose point x on qr so that ab = qx.
2. draw \\(xy\parallel rs\\) such that y lies on \\(\overline{qs}\\).

reasons

1. construction

2.

Explanation:

Step1: State given info

$\frac{AB}{QR} = \frac{AC}{QS}$, $\angle A \cong \angle Q$

Step2: Construct segment XY

Choose $X$ on $\overline{QR}$: $AB = QX$; draw $\overline{XY} \parallel \overline{RS}$, $Y$ on $\overline{QS}$

Step3: Prove $\Delta QXY \sim \Delta QRS$

By Basic Proportionality Theorem (AA similarity: $\angle Q$ common, $\angle QXY \cong \angle QRS$, $\angle QYX \cong \angle QSR$), so $\frac{QX}{QR} = \frac{QY}{QS}$

Step4: Substitute QX = AB

$\frac{AB}{QR} = \frac{QY}{QS}$. Given $\frac{AB}{QR} = \frac{AC}{QS}$, so $QY = AC$

Step5: Prove $\Delta ABC \cong \Delta QXY$

By SAS congruence: $AB = QX$, $\angle A \cong \angle Q$, $AC = QY$

Step6: Link congruence to similarity

Since $\Delta ABC \cong \Delta QXY$ and $\Delta QXY \sim \Delta QRS$, then $\Delta ABC \sim \Delta QRS$

Answer:

$\Delta ABC \sim \Delta QRS$ is proven via construction, similarity of triangles, congruence, and transitivity of similarity.

Completed Statement-Reason Table:

Statements	Reasons
2. Draw $\overline{XY} \parallel \overline{RS}$ such that $Y$ lies on $\overline{QS}$	2. Construction (Parallel Postulate)
3. $\Delta QXY \sim \Delta QRS$	3. AA Similarity Postulate
4. $\frac{QX}{QR} = \frac{QY}{QS}$	4. Corresponding sides of similar triangles are proportional
5. $\frac{AB}{QR} = \frac{QY}{QS}$	5. Substitution ($AB = QX$)
6. $\frac{AB}{QR} = \frac{AC}{QS}$	6. Given
7. $QY = AC$	7. Transitive property of equal ratios
8. $\Delta ABC \cong \Delta QXY$	8. SAS Congruence Postulate
9. $\Delta ABC \sim \Delta QRS$	9. Transitivity of similarity (if $\cong$ then $\sim$, and $\Delta QXY \sim \Delta QRS$)