Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Proof:
Let's consider a triangle ABC, where DE is drawn parallel to side BC, intersecting AB at point D and AC at point E.

Step 1: Proving the proportionality of AD and DB.
Since DE is parallel to BC, we can apply the intercept theorem to triangles ADE and ABC. According to the intercept theorem, the following proportionality holds:

AD/DB = AE/EC

Step 2: Proving the proportionality of AE and EC.
Now, let's consider triangle ADE. Since DE is parallel to BC, we can apply the intercept theorem to triangles ADE and ABC. According to the intercept theorem, the following proportionality holds:

AE/EC = AD/DB

Step 3: Combining the results.
From steps 1 and 2, we have:
AD/DB = AE/EC
AE/EC = AD/DB

Therefore, we can conclude that the other two sides, AB and AC, are divided in the same ratio by the line DE.

Proof that triangle ABC is an isosceles triangle if DE is parallel to BC and BD = CE:
Given that DE is parallel to BC and BD = CE, we need to prove that triangle ABC is an isosceles triangle.

Step 1: Proving the proportionality of AD and DB.
Using the given information, we have BD = CE. From the theorem proved above, we know that if a line is drawn parallel to one side of a triangle and intersects the other two sides in distinct points, the other two sides are divided in the same ratio. Therefore, we can conclude that AD/DB = AE/EC.

Step 2: Proving AB = AC.
Since AD/DB = AE/EC and BD = CE, we can substitute these values into the proportionality equation:
AD/BD = AE/CE

Since AD = AE (as they are two sides of the same line segment DE), and BD = CE (given), the equation becomes:
AD/BD = AD/CE

Cancelling out AD from both sides, we get:
1/BD = 1/CE

Cross multiplying, we have:
CE = BD

Since BD = CE, we can conclude that AB = AC.

Therefore, triangle ABC is an isosceles triangle, as desired.