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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
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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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Construct triangle BCP, when CB = 5 cm, BP = 4 cm, ∠PBC = 45°.
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Ruler and compass only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
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(ii) Construct the locus of point 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 as P, the point which is equidistant from AB, BC and also equidistant from B and C.
(v) Measure and record the length of PB.
