Question

Given: $P$is on the perpendicular bisector of $\overline{AB}$; $P$is on the perpendicular bisector of $\overline{BC}$.

Prove: $PA=PC$

Short answer

It is being given that $P$is on the perpendicular bisector of $\overline{AB}$, therefore by using theorem 4-5 it can be noticed that the point $P$is equidistant from the endpoints of the segment role="math" localid="1649157167701" $\overline{AB}$.

that implies, $PA=PB$.

It is also being given that $P$is on the perpendicular bisector of $\overline{BC}$, therefore by using theorem 4-5 it can be noticed that the point $P$is equidistant from the endpoints of the segment $\overline{BC}$.

that implies, $PB=PC$.

As, $PA=PB$and $PB=PC$, therefore it can be noticed that.

Hence proved.

Step-by-step solution

Step 1. Write the definition of perpendicular bisector.

The perpendicular bisector to a line is the line that is perpendicular to the given line and also passes through the midpoint of the given line.

Step 2. Write the theorem 4-5.

Theorem 4-5 states that if a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.

Step 3. Write the proof of the statement that PA=PC.

It is being given that $P$is on the perpendicular bisector of $\overline{AB}$, therefore by using theorem 4-5 it can be noticed that the point $P$is equidistant from the endpoints of the segment $\overline{AB}$.

that implies, $PA=PB$.

It is also being given that $P$is on the perpendicular bisector of $\overline{BC}$, therefore by using theorem 4-5 it can be noticed that the point$P$ is equidistant from the endpoints of the segment $\overline{BC}$.

that implies, $PB=PC$.

As,$PA=PB$ and $PB=PC$, therefore it can be noticed that $PA=PC$.

Hence proved.