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प्रश्न
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
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उत्तर
Steps of construction:
- Draw a ray BC.
- Construct a ray RA making an angle of 75° with BC. Therefore, ABC = ABC = 75°
- Draw the angle bisector BP of ∠ABC.
BP is the required locus. - Take any point D on BP.
- From D, draw DE ⊥ AB and DF ⊥ BC.
Since D lies on the angle bisector BP of ∠ABC.
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.
संबंधित प्रश्न
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
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.
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