Question

given: \\( \overline { d f } \\) and \\( \overline { d e } \\) are midsegments of \\( \triangle a b c \\) prove: \\( d e = \frac { 1 } { 2 } a c \\) what is the missing step in this proof? scroll down to see the entire proof. a. statement: \\( d e = \frac { 1 } { 2 } a f \\) reason: definition of midpoint b. statement: \\( d e = a f \\) reason: definition of midpoint c. statement: \\( \overline { d e } \\) is parallel to \\( \overline { a c } \\) reason: definition of parallel lines d. statement: \\( d e = a f \\) reason: corresponding sides of congruent triangles are congruent

Explanation:

Step1: Analyze the proof structure

We have \(\triangle DBE\cong\triangle ADF\) (by ASA in step 6).

Step2: Use the property of congruent triangles

When two triangles are congruent, their corresponding sides are congruent. So, if \(\triangle DBE\cong\triangle ADF\), then \(DE = AF\) (by the property that corresponding sides of congruent triangles are congruent).

Answer:

D. Statement: \(DE = AF\) Reason: Corresponding sides of congruent triangles are congruent.