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.

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

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

Proof:

 Statement1. ∠YXP = ∠XQZ2. ∠PXZ = ∠XZQ3. ∠XQZ = ∠XZQ4. XQ = XZ5. $$\frac{YX}{XQ}$$ = $$\frac{YP}{PZ}$$ 6. $$\frac{YX}{XZ}$$ = $$\frac{YP}{PZ}$$ Reason1. XP  ∥ QZ and YQ is a transversal2. XP  ∥ QZ and XZ is a transversal3. ∠YXP = ∠PXZ4. ∠XQZ = ∠XZQ5. 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.

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