Question

in triangle abc, \\(\frac{da}{bd} = \frac{ec}{be}\\).
move options to the blanks to prove that \\(\overline{de} \parallel \overline{ac}\\).
it is given that \\(\frac{da}{bd} = \frac{ec}{be}\\). by the addition property of equality, \\(\frac{bd}{bd}+\underline{quad} = \frac{be}{be}+\underline{quad}\\). this equation can be simplified to \\(\frac{bd + da}{bd} = \frac{be + ec}{be}\\). then, by the segment addition postulate, \\(\underline{quad} = \underline{quad}\\). additionally, \\(\underline{quad} \cong \underline{quad}\\) by the reflexive property of congruence. as a result, \\(\triangle abc \sim \triangle dbe\\) by side - angle - side similarity. because corresponding angles of similar triangles are congruent, \\(\angle 1 \cong \underline{quad}\\). finally, \\(\overline{de} \parallel \overline{ac}\\) because the corresponding angles of parallel lines are congruent.
options: \\(\frac{ba}{bd}\\), \\(\frac{bc}{be}\\), \\(\frac{da}{bd}\\), \\(\frac{ec}{be}\\), \\(\angle a\\), \\(\angle b\\), \\(\angle c\\), \\(\angle 1\\), \\(\angle 2\\)

Explanation:

1. To apply the Addition Property of Equality to the given proportion, we add the original fractions to both sides of the equation.
2. Using the Segment Addition Postulate, we recognize that the sum of segments makes up the full sides of the triangle.
3. The Reflexive Property of Congruence tells us an angle is congruent to itself, which is the shared angle here.
4. Corresponding angles of similar triangles are congruent, so we match the corresponding angle from the larger triangle to $\angle 1$.

Answer:

It is given that $\frac{DA}{BD} = \frac{EC}{BE}$. By the Addition Property of Equality, $\frac{BD}{BD}+\boldsymbol{\frac{DA}{BD}} = \frac{BE}{BE}+\boldsymbol{\frac{EC}{BE}}$. This equation can be simplified to $\frac{BD+DA}{BD} = \frac{BE+EC}{BE}$. Then, by the Segment Addition Postulate, $\boldsymbol{\frac{BA}{BD}} = \boldsymbol{\frac{BC}{BE}}$. Additionally, $\boldsymbol{\angle B} \cong \boldsymbol{\angle B}$ by the Reflexive Property of Congruence. As a result, $\triangle ABC \sim \triangle DBE$ by Side-Angle-Side similarity. Because corresponding angles of similar triangles are congruent, $\angle 1 \cong \boldsymbol{\angle A}$. Finally, $\overline{DE} \parallel \overline{AC}$ because the corresponding angles of parallel lines are congruent.