Draw a straight line AB of length 8cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.

We know that locus of point equidistant from two points is the perpendicular bisector of the line segment joining them.

From figure,

CD is the locus of all points which are equidistant from A and B.

Proof :

Consider △GOA and △GOB.

∠GOA = ∠GOB (Both are equal to 90°)

OG = OG (Common side)

AO = OB (They are equal as CD bisects AB at O).

Hence, by SAS axiom △GOA ~ △GOB.

Since triangles are similar, hence the ratio of their corresponding sides are equal.

$\therefore \dfrac{AG}{OG} = \dfrac{BG}{OG} \\[1em] \Rightarrow AG = \dfrac{BC}{OG} \times OG \\[1em] \Rightarrow AG = BG.$

Hence, proved that AG = BG. Thus proved that any point on CD is equidistant from A and B.