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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

Δ BMC ≅ Δ CNB...

Question

$Δ B M C ≅ Δ C N B$
If the above statement is true then mention answer as 1, else mention 0 if false

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Solution

In $△$ s, BMC and BNC,
$B C = B C$ (Common side)
$B N = C M$ (Given)
$∠ B = ∠ C$ (By Isosceles triangle property, since, $A B = A C$ )
Thus, $△ B M C ≅ △ B N C$ (SAS congruency)

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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector

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