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. 

Answer 1

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.