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Question
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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Solution
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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