Question

ABC is a triangle . Locate a point in the interior of triangle ABC which is equaidistant from all the vertices of triangle ABC

Answer 1

Answer:

Let ABC be a triangle .

draw a ptependicular bidectors PQandRS of sides ABand BC respectively of triangle ABC. let PQbisects AB at M and RS bisects BC at point N.

Let PQ and RS intersect at point O.

Join OA, OB, OC.

Now in triangle AOM and triangle BOM,

AB=MB [by construction]

angle AMO=angle BMO=90°[ by construction]

OM=OM[common]

therefore triangle oem is congruent to triangle BOM [By SAS congruency]

:OA= OB[by C.P.C.T.]......(1)

similarly, triangle BON is congruent to triangle CON

:OB =OC [NYC.P.C.T.]......(2)

From eq.(1) and (2)

OA =OB =OC

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