Application of Basic Proportionality Theorem

Here we will prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.

Given: XP is the internal bisector of ∠YXZ, intersecting YZ at P.

Application of Basic Proportionality Theorem

To prove: \(\frac{YP}{PZ}\) = \(\frac{XY}{XZ}\).

Construction: Draw ZQ  ∥ XP such that ZQ meets YX produced at Q.



1. ∠YXP = ∠XQZ

2. ∠PXZ = ∠XZQ

3. ∠XQZ = ∠XZQ

4. XQ = XZ

5. \(\frac{YX}{XQ}\) = \(\frac{YP}{PZ}\)

6. \(\frac{YX}{XZ}\) = \(\frac{YP}{PZ}\)


1. XP  ∥ QZ and YQ is a transversal

2. XP  ∥ QZ and XZ is a transversal

3. ∠YXP = ∠PXZ

4. ∠XQZ = ∠XZQ

5. XP ∥ QZ

6. By statement 4.


1. The above proposition is true for external division also.

So, \(\frac{YP}{ZP}\) = \(\frac{XY}{XZ}\)

Application of Basic Proportionality Theorem Image

2. Converse of the above proposition is also true.

So, if P is a point on YZ such that YP : PZ = XY : XZ then XP bisects the angle YXZ internally or externally.

9th Grade Math

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