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 \(\frac{XP}{PY}\) = \(\frac{XQ}{QZ}\).

Converse of Basic Proportionality Theorem

To prove: PQ ∥ YZ




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


(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.

9th Grade Math

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