Question

in the figure, line / divides the side lengths of \\( \triangle abc \\) proportionally such that \\( \frac{db}{ad}=\frac{ec}{ae} \\).
move options to the proof to show that line / is parallel to \\( \overline{bc} \\).

statements	reasons
\\( \frac{db}{ad}\square\frac{ad}{ad}=\frac{ec}{ae}\square\frac{ae}{ae} \\)	property of equality
\\( \frac{ab}{ad}=\frac{ac}{ae} \\)	substitution
\\( \angle a \cong \angle a \\)	reflexive property
\\( \triangle abc\square\triangle ade \\)	side - angle - side
\\( \angle abc\square\angle ade \\)	corresponding angles of triangles are
line / is parallel to \\( \overline{bc} \\)	corresponding angles theorem

Explanation:

Step1: Add 1 to both sides

$$\frac{DB}{AD}+1=\frac{EC}{AE}+1$$

Step2: Use segment addition

Since \(AB = AD + DB\) and \(AC=AE + EC\), we substitute into the equation from Step 1.
$$\frac{AD + DB}{AD}=\frac{AE + EC}{AE}$$
Which simplifies to \(\frac{AB}{AD}=\frac{AC}{AE}\)

Step3: Use similarity criterion

For \(\triangle ABC\) and \(\triangle ADE\), we have \(\frac{AB}{AD}=\frac{AC}{AE}\) (from above) and \(\angle A\cong\angle A\) (reflexive property). By the Side - Angle - Side (SAS) similarity criterion, \(\triangle ABC\sim\triangle ADE\)

Step4: Use similarity property

Corresponding angles of similar triangles are congruent. So, \(\angle ABC\cong\angle ADE\)

Answer:

For the first blank (operation on \(\frac{DB}{AD}=\frac{EC}{AE}\)): \(+\), \(+\) (using the Addition Property of Equality). For the second blank (\(\triangle ABC\) and \(\triangle ADE\)): \(\sim\). For the third blank (\(\angle ABC\) and \(\angle ADE\)): \(\cong\), similar, congruent.