Question

in triangle abc, $\frac{da}{bd}=\frac{ec}{be}$. move options to the blanks to prove that $overline{de}paralleloverline{ac}$. it is given that $\frac{da}{bd}=\frac{ec}{be}$. by the addition property of equality, $\frac{bd}{bd}+$__$=\frac{be}{be}+$. this equation can be simplified to $\frac{bd + da}{bd}=\frac{be+ec}{be}$. then, by the segment addition postulate, = . additionally, $cong$ by the reflexive property of congruence. as a result, $\triangle abcsim\triangle dbe$ by side - angle - side similarity. because corresponding angles of similar triangles are congruent, $angle1cong$ __. finally, $overline{de}paralleloverline{ac}$ because the corresponding angles of parallel lines are congruent. $\frac{ba}{bd}$, $\frac{bc}{be}$, $\frac{da}{bd}$, $\frac{ec}{be}$, $angle a$, $angle b$, $angle c$, $angle1$, $angle2$

Explanation:

Step1: Apply addition property

$\frac{BD}{BD}+\frac{DA}{BD}=\frac{BE}{BE}+\frac{EC}{BE}$

Step2: Segment addition postulate

$\frac{BA}{BD}=\frac{BC}{BE}$

Step3: Reflexive property

$\angle B\cong\angle B$

Step4: Similar - triangle property

$\angle1\cong\angle A$

Answer:

First blank: $\frac{DA}{BD}$; Second blank: $\frac{EC}{BE}$; Third blank: $\frac{BA}{BD}$; Fourth blank: $\frac{BC}{BE}$; Fifth blank: $\angle B$; Sixth blank: $\angle B$; Seventh blank: $\angle A$