Question

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. Here XY is the _____

• A. perpendicular bisector of XY
• B. perpendicular but not the bisector of XY
• C. only the bisector of XY
• D. perpendicular bisector of AX

Answer 1

The correct option is A perpendicular bisector of XY

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

Join AX, AY and BX, BY.
Consider $△$ AXY and $△$ BXY,

AX = AY = BX = BY (same radius)
XY = XY (common side)

$△$ AXY $\cong$ $△$ BXY (By S.S.S congruency)
$\therefore$ $\angle$ AXY = $\angle$ BXY (by CPCT)
Now consider $△$ AXO and $△$ BXO:
$△$ AXO $\cong$ $△$ BXO
by SAS,
as AX = BX ( same radius)
OX is the common side and
$\angle$ AXY = $\angle$ BXY ( proved above)
$\therefore$ $△$ AXO $\cong$ $△$ BXO
$\Rightarrow$ AO = BO
which means XY bisects AB.
Now, $\angle$ AOB = ${180}^{\circ }$
$\Rightarrow$ $\angle$ AOX = $\angle$ BOX = ${90}^{\circ }$
$\Rightarrow$ XY is the perpendicular bisector of AB