In triangle ABC, the construction required to find point P equidistant from AB and AC and also, equidistant from B and C is/are
In triangle ABC, the construction required to find point P equidistant from AB and AC and also, equidistant from B and C is/are
The correct options are
A Bisector of angle A
B Perpendicular bisector of side BC
The theorem, locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines, suggests that point P should be on the angular bisector of angle A to be equidistant from sides AB and AC.
Also, the theorem, locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points, concludes that the point P must lie on the perpendicular bisector of BC to be equidistant from B and C.
Hence, the point P is at the intersection of the angular bisector of A and perpendicular bisector of BC.
A
Bisector of angle A
B
Perpendicular bisector of side BC
The theorem, locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines, suggests that point P should be on the angular bisector of angle A to be equidistant from sides AB and AC.
Also, the theorem, locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points, concludes that the point P must lie on the perpendicular bisector of BC to be equidistant from B and C.
Hence, the point P is at the intersection of the angular bisector of A and perpendicular bisector of BC.
