Question

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

Answer 1

Answer:

ABC is a triangle. Locate a point in the interior of ABC which is equidistant from all the vertices of ABC. Ans. ... Hence O, the point of intersection of perpendicular bisectors of any two sides of ABC equidistant from its vertices.

Answer 2

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.