Proof:

To prove that a line drawn through the midpoint of one side of a triangle, parallel to the other side, bisects the third side, we can use the concept of similar triangles.

Given: Triangle ABC, with side AB, side AC, and side BC. Let D be the midpoint of side AB, and let line DE be drawn parallel to side AC, intersecting side BC at point E.

To Prove: Line DE bisects side BC.

Proof:

Step 1: Establish a relationship between triangles ADE and ABC.

Since line DE is parallel to side AC, we can conclude that angles AED and BAC are corresponding angles and therefore congruent. Additionally, since DE is parallel to AC, we have angle DEA congruent to angle CBA (alternate interior angles).

Step 2: Show that triangles AED and ABC are similar.

Using the Angle-Angle (AA) similarity criterion, we have established that triangles ADE and ABC are similar. This means that the corresponding sides are proportional.

Step 3: Show that side DE is proportional to side BC.

Since triangles ADE and ABC are similar, we can write the following ratio:

DE/AC = AD/AB

Since D is the midpoint of AB, we know that AD = DB. Therefore, we can rewrite the ratio as:

DE/AC = DB/AB

But since DE is parallel to AC, we know that DE = AC/2. Substituting this information into the ratio, we have:

AC/2/AC = DB/AB

Simplifying the ratio further, we get:

1/2 = DB/AB

Step 4: Conclude that line DE bisects side BC.

From the previous step, we established that DB/AB = 1/2. This means that line DE divides side BC into two segments, BE and EC, in the ratio of 1:2. Therefore, line DE bisects side BC.

Conclusion:

Based on the above proof, we can conclude that a line drawn through the midpoint of one side of a triangle, parallel to the other side, bisects the third side.