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.

Proof:

Statement

1. ∠YXP = ∠XQZ

2. ∠PXZ = ∠XZQ

3. ∠XQZ = ∠XZQ

4. XQ = XZ

5. $\frac{YX}{XQ}$ = $\frac{YP}{PZ}$

6. $\frac{YX}{XZ}$ = $\frac{YP}{PZ}$

Reason

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, $\frac{YP}{ZP}$ = $\frac{XY}{XZ}$

$Application of Basic Proportionality Theorem Image$

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.

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