# Converse of Basic Proportionality Theorem

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.