मराठी

In Liabc, the Perpendicular Bisector of Ab and Ac Meet at 0. Prove that 0 is Equidistant from the Three Vertices

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प्रश्न

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. 

बेरीज
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उत्तर

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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पाठ 15: Loci - Exercise 16.1

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संबंधित प्रश्‍न

In each of the given figures; PA = PB and QA = QB. 

i.
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Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.


Use ruler and compasses only for this question.

  1. Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
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  4. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.

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