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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:
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Statement |
Reason |
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1. \(\frac{XP}{PY}\) = \(\frac{XQ}{QZ}\). |
1. Given |
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2. \(\frac{PY}{XP}\) = \(\frac{QZ}{XQ}\) |
2. Taking reciprocals of both sides in statement 1. |
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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. |
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4. In ∆XYZ and ∆XPQ, (i) \(\frac{XY}{XP}\) = \(\frac{XZ}{XQ}\) (ii) ∠YXZ = ∠PXQ |
4. (i) From statement 3. (ii) Common angle |
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5. Therefore, ∆XYZ ∼ ∆XPQ |
5. By SAS criterion of similarity. |
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6. Therefore, ∠XYZ = ∠XPQ |
6. Corresponding angles of similar triangles are equal. |
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7. YZ ∥ PQ |
7. Corresponding angles are equal. |
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