Math, asked by Asifkamal55, 3 months ago

1. ABC is a triangle. Locate a point in the interior of A ABC which is equidistant from all
the vertices of A ABC.
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Answers

Answered by mdmonazirhussain

Answer:Let OD and OE be the perpendicular bisectors of sides BC and AB of △ABC respectively.

∴ By perpendicular bisector theorem,

O is equidistant from the end points of seg BC i.e. points B and C.

Similarly, point O is equidistant from end points of seg AC i.e points C and A.

Hence, the point of intersection O of the perpendicular bisectors of sides AB and BC is equidistant from vertices A,B,C of △ABC.

Step-by-step explanation:

Answered by ButterflyUSR

Answer:

Answer:Let OD and OE be the perpendicular bisectors of sides BC and AB of △ABC respectively.

∴ By perpendicular bisector theorem,

O is equidistant from the end points of seg BC i.e. points B and C.

Similarly, point O is equidistant from end points of seg AC i.e points C and A.

Hence, the point of intersection O of the perpendicular bisectors of sides AB and BC is equidistant from vertices A,B,C of △ABC.

BRO I THANKED ALL YOUR ANSWERS PLEASE GIVE ONE BRAINLIEST

Step-by-step explanation:

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