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
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:
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.
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Step-by-step explanation: