Question
Each point on a/an _________ is such that it forms an isosceles triangle with the end points of the given line segment.
Answer: Option A
:
A
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∘
Also PM is common side
So △ PMA ≅ △ PMB (SASRule)
∴ PA = PB (CPCT)
Hence, △ PAB is an isosceles triangle.
So, the given statement is true.
Was this answer helpful ?
:
A
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∘
Also PM is common side
So △ PMA ≅ △ PMB (SASRule)
∴ PA = PB (CPCT)
Hence, △ PAB is an isosceles triangle.
So, the given statement is true.
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution