Question

Use the diagrams on pages 153 and 154 to prove the following theorems.

Theorem 4-7

Short answer

The diagram is:

From the diagram, it can be noticed that$PB$is the angle bisector of the angle $\angle B$and the point$p$is lying on the angle bisector.

In the triangles role="math" localid="1649230725652" $△PXB$and $△PYB$, it can be noticed that:

$$\begin{gathered} \angle XBP=\angle YBP(\because PB\mathrm{is}\mathrm{angle}\mathrm{bisector}) \\ \angle BXP=\angle BYP=90° \\ BP\cong BP\left(\mathrm{common}\right) \end{gathered}$$

Therefore, it can be noticed that $\angle XBP\cong \angle YBP$,$\angle BXP\cong \angle BYP$and $BP\cong BP$.

Therefore, the triangles $△PXB$and $△PYB$are congruent triangles by using the AAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $PX\cong PY$.

Therefore the pointis equidistant from the sides of the angle.

Therefore, if a point lies on the bisector of an angle then the point is equidistant from the sides of the angle.

Therefore, Theorem 4-7 is proved.

Step-by-step solution

Step 1. Write the definition of an angle bisector.

The angle bisector is a ray that divides the angle into two equal angles.

Step 2. Write the theorem 4-7.

The theorem 4-7 states that if a point lies on the bisector of an angle then the point is equidistant from the sides of the angle.

Step 3. Draw the diagram to prove theorem 4-7.

The diagram is:

From the diagram, it can be noticed that $PB$ is the angle bisector of the angle $\angle B$and the point $p$ is lying on the angle bisector.

Step 4. Write the proof of the theorem 4-7.

In thetriangles $△PXB$and $△PYB$, it can be noticed that:

role="math" localid="1649231285910" $\angle XBP=\angle YBP(\because PB\mathbf{is}\mathbf{}\mathbf{angle}\mathbf{}\mathbf{bisector})$

$\angle BXP=\angle BYP=\mathbf{90}\mathbf{°}$

$BP\cong BP\left(\mathrm{common}\right)$

Therefore, it can be noticed that $\angle XBP\cong \angle YBP$,$\angle BXP\cong \angle BYP$and $BP\cong BP$.

Therefore, the triangles$△PXB$and$△PYB$are congruent triangles by using the AAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $PX\cong PY$.

Therefore the point$P$ is equidistant from the sides of the angle.

Therefore, if a point lies on the bisector of an angle then the point is equidistant from the sides of the angle.

Therefore, Theorem 4-7 is proved.