Question

Each point on a/an _________ is such that it forms an isosceles triangle with the end points of the given line segment.

• A. perpendicular bisector
• B. angular bisector
• C. altitude
• D. median

Answer 1

The correct option is A perpendicular bisector

Consider the above figure.
Here, XY is the perpendicular bisector of a line AB.
Let P be any random point on XY.
In $△$ PMA $\cong$ $△$ PMB
AM = BM (Perpendicular bisector divides a line segment into two equal halves)
$\angle$ PMA = ​$\angle$ PMB = 90${}^{\circ }$
Also PM is common side
So $△$ PMA $\cong$ $△$ PMB (SAS Rule)
$\therefore$ PA = PB (CPCT)
Hence, $△$ PAB is an isosceles triangle.
So, the given statement is true.