Prove that the right bisectors of sides of a triangle are concurrent.
Asked by hermionee | 24th Apr, 2009, 05:37: PM
Consider a triangle ABC.
Since AB and AC are intersecting line segments,so we can say that their right bisectors must also intersect.
Let them meet in P (say)
We know that every point the right bisector of a segment is equidistant from the end points.
So,
AP=PC (i)...(P lies on the right bisector of AB)
and
AP=PC..(ii) (as P lies on the right bisector of AC)
from (i) and (ii)
BP=PC
Thus, we can say that P lies on the right bisector of BC, as it's equidistant from B and C.
Thus we see that P lies on the right bisectors of all therr sides of the triangle.
Thus , proved.
This question can be done by taking any of the two sides of the triangle in the beginning. Method remains the same.
Answered by | 29th Apr, 2009, 04:01: PM
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