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प्रश्न
In the given figure ABC is a triangle. CP bisects angle ACB and MN is perpendicular bisector of BC. MN cuts CP at Q. Prove Q is equidistant from B and C, and also that Q is equidistant from BC and AC.
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उत्तर
Join BQ and draw perpendicular bisector of AC cutting AC at L.
In Δ QBN and ΔQCN
QN = QN
BN =NC
∠ QNB = ∠ QNC = 90 degree.
Therefore, ∠ QBN and ∠.QCN are congruent .
Hence Q is equidistant from B and C.
In Δ QNC and Δ QLC
QC= QC
∠ QLC = ∠ QNC = 90 degree.
∠ QCL =∠ QCN (PC being angle bisector)
Therefore, .Δ QNC and Δ QLC are congruent.
Therefore, QL = QN.
Hence Q is equidistant from BC and AC.
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संबंधित प्रश्न
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°
Hence:
1) Construct the locus of points equidistant from BA and BC
2) Construct the locus of points equidistant from B and C.
3) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Use ruler and compasses only for this question:
I. Construct ABC, where AB = 3.5 cm, BC = 6 cm and ABC = 60o.
II. Construct the locus of points inside the triangle which are equidistant from BA and BC.
III. Construct the locus of points inside the triangle which are equidistant from B and C.
IV. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and records the length of PB.
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- equidistant from BA and BC.
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- 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
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(ii) the locus of point equidistant from A and C.
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(i) Mark two points D and E which satisfy the condition that they are equidistant from both ABA and BC.
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