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.

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

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

**Proof:**

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. |

**Note:**

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

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

**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.

**From Application of Basic Proportionality Theorem to HOME PAGE**

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