Question

AB is a line segment. With A and B as centres and any radius greater than half of AB, draw arcs on either side of AB so that they meet at X and Y as shown. Join XY

There are the two statements:
Statement 1: AB is the perpendicular bisector of XY
Statement 2: XY is the perpendicular bisector of AB
Choose the correct option from below.

• A. Only Statement 1 is true
• B. Only Statement 2 is true
• C. Both the statements are true
• D. None of the above

Answer 1

The correct option is C Both the statements are true

If AB is a perpendicular bisector of line XY, it should divide XY in such a way that XO = OY and $\angle$ AOX = $\angle$ AOY = ${90}^{\circ }$.
We shall see if it is true or not.
Since XY is the perpendicular bisector, AO = OB.

Join AX, AY and BX, BY.
Consider $△$ AXY and $△$ BXY,
AX = AY = BX = BY (same radius)
XY = XY (common side)
$\therefore$ $△$ AXY $\cong$ $△$ BXY (By S.S.S congruency)

$\angle$ AOX = $\angle$ BOX = ${90}^{\circ }$ (since $\angle$ AOB = ${180}^{\circ }$).
Similarly, $\angle$ AOY = $\angle$ BOY = ${90}^{\circ }$.

Now, in $△$ AOX and $△$ BOY,
AX = BY,
AO = OB,
$\angle$ AOX = $\angle$ BOY = ${90}^{\circ }$,
$\therefore$ $△$ AOX $\cong$ $△$ BOY (RHS postulate)
Hence OX = OY.
So both the statements are true.