Question

ABC is a triangle. Locate a point in the interior of ANC which is equidistant from all the vertices of ABC. ​

Answer 1

Step-by-step explanation:

Let ABC be a triangle.

Draw perpendicular bisectors PQ and RS of sides AB and BC respectively of triangle ABC. Let PQ bisects AB at M and RS bisects BC at point N.

Let PQ and RS intersect at point O.

Join OA, OB and OC.

Now in AOM and BOM,

AM = MB [By construction]

AMO = BMO = [By construction]

OM = OM [Common]

AOM BOM [By SAS congruency]

OA = OB [By C.P.C.T.] …..(i)

Similarly, BON CON

OB = OC [By C.P.C.T.] …..(ii)

From eq. (i) and (ii),

OA = OB = OC

Hence O, the point of intersection of perpendicular bisectors of any two sides of ABC equidistant from its vertices.

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Answer 2

Answer:

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