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प्रश्न
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
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उत्तर
Steps of construction:
- In the given triangle, draw the angle bisector of ∠BAC.
- Draw the perpendicular bisector of BC which intersects the angle bisector at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.
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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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- 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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