Here we will prove converse of basic proportionality theorem.
The line dividing two sides of a triangle proportionally is parallel to the third side.
Given: In ∆XYZ, P and Q are points on XY and XZ respectively, such that \(\frac{XP}{PY}\) = \(\frac{XQ}{QZ}\).
To prove: PQ ∥ YZ
Proof:
Statement 
Reason 
1. \(\frac{XP}{PY}\) = \(\frac{XQ}{QZ}\). 
1. Given 
2. \(\frac{PY}{XP}\) = \(\frac{QZ}{XQ}\) 
2. Taking reciprocals of both sides in statement 1. 
3. \(\frac{PY}{XP}\) + 1 = \(\frac{QZ}{XQ}\) + 1 ⟹ \(\frac{PY + XP}{XP}\) = \(\frac{QZ + XQ}{XQ}\) ⟹ \(\frac{XY}{XP}\) = \(\frac{XZ}{XQ}\) 
3. By adding 1 on both sides of statement 2. 
4. In ∆XYZ and ∆XPQ, (i) \(\frac{XY}{XP}\) = \(\frac{XZ}{XQ}\) (ii) ∠YXZ = ∠PXQ 
4. (i) From statement 3. (ii) Common angle 
5. Therefore, ∆XYZ ∼ ∆XPQ 
5. By SAS criterion of similarity. 
6. Therefore, ∠XYZ = ∠XPQ 
6. Corresponding angles of similar triangles are equal. 
7. YZ ∥ PQ 
7. Corresponding angles are equal. 
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