**Converse of BPT:**

The Converse of the Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally.

**Proof of the Converse of BPT:**

Let's consider a triangle ABC and a line DE parallel to side BC, where D lies on AB and E lies on AC.

**1. Assume DE divides sides AB and AC proportionally:**

Let the lengths of the segments be AD = x, DB = y, AE = m, and EC = n. According to the assumption, DE divides sides AB and AC proportionally, so we can write:

AD/DB = AE/EC = x/y = m/n

**2. Assume DE intersects side BC at point F:**

Let the length of segment BF be p and FC be q. Therefore, we have:

DB = BF + p
EC = FC + q

**3. Apply BPT to triangle ABC and line DE:**

According to the Basic Proportionality Theorem, if a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. Applying this theorem to triangle ABC and line DE, we get:

x/y = m/n = (BF + p)/(FC + q)

**4. Prove that DE is parallel to BC:**

To prove that DE is parallel to BC, we need to show that the alternate interior angles are equal. Since line DE is parallel to BC, we have:

∠DEB = ∠ABC (corresponding angles)
∠DEC = ∠ACB (corresponding angles)

**5. Use similar triangles to prove proportionality:**

Now, we can use similar triangles to prove that the lengths are proportional. Triangles DBF and ECF are similar (AAA similarity) because:

∠DBF = ∠ECF (corresponding angles)
∠BDF = ∠CEF (alternate interior angles)
∠BFD = ∠CFE (corresponding angles)

Therefore, we can write:

BF/FC = DB/EC = y/n

**6. Conclude that DE divides AB and AC proportionally:**

From steps 3 and 5, we have:

x/y = m/n = (BF + p)/(FC + q) = BF/FC = y/n

Therefore, DE divides sides AB and AC proportionally, which proves the converse of the Basic Proportionality Theorem.