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Question
In Δ ABC, the perpendicular bisector of AB and AC meet at 0. Prove that O is equidistant from the three vertices. Also, prove that if M is the mid-point of BC then OM meets BC at right angles.
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Solution
Since O lies on the perpendirular bisector of AB, O is equidistant from A and B.
OA = OB ........ (i)
Again, O lies on the perpendirular bisector of AC, O is equidistant from A and C.
OA = OC ......... (ii)
From (i) and (ii)
OB= OC
Now in Δ OBM and Δ OCM,
OB = OC (proved)
OM=OM
BM =CM ( M is mid-point of BC)
Therefore, Δ OBM and Δ OCM are congruent.
∠ OMB= ∠ OMC
But BMC is a straight line, so
∠ OMB =∠ OMC = 90°
Thus, OM meets BC at right angles.
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