Question

Draw any angle with vertex $O$. Take a point $A$ on one of its arms and $B$ on another such that $OA=OB$. Draw the perpendicular bisectors of $OA$ and $OB$. Let them meet at $P$. Is $PA=PB$?

Answer 1

Draw the diagram with the help of a given statement.

1. Draw any angle with $O$ as the vertex.

2. Draw arcs on both rays of this angle using $O$ as the center and an appropriate radius. Let's these two points be $A$ and $B$.

3. Draw arcs on both sides of $OA$, using $O$ and $A$ as centers and a radius greater than half of $OA$. Allow these to intersect at $C$ and $D$, respectively. Join the $CD$.

4. Draw two arcs on each side of $OB$ using $O$ and $B$ as centers and a radius of more than half of $OB$. Label the points where the arcs intersect as $E$ and $F$.

Construct a perpendicular bisector of $OB$ by joining $EF$.

At point $P$, the perpendicular bisectors $CD$ and $EF$ intersect. Then, measure the length of $PA$ and $PB$ by compass, note that their lengths are equal.

Yes, $PA=PB$.

Hene, the diagram is as follows: