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

perpendicular bisector

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angular bisector

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altitude

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median

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Solution

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 $≅$ $△$ PMB
AM = BM (Perpendicular bisector divides a line segment into two equal halves)
$∠$ PMA = $∠$ PMB = 90 $^{\circ}$
Also PM is common side
So $△$ PMA $≅$ $△$ PMB (SAS Rule)
$∴$ PA = PB (CPCT)
Hence, $△$ PAB is an isosceles triangle.
So, the given statement is true.

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

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