Construct a triangle ABC, in which $A B C = 75^{o}$ ,AB=5 cm and BC=6.4 cm. Draw perpendicular bisector of side BC and also the bisector of Angle ACB.If these bisectors intersect each other at point P;prove that P is equidistant from B and C; and also from AC and BC.

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Solution

Steps of Construction:
i)Draw a line segment BC= 6.4 cm
ii)At B, draw a ray BX making an angle of75°with BC and cut off BA = 5 cm.
iii)Join AC.
is the required triangle.
iv) Draw the perpendicular bisector of BC.
v) Draw the angle bisector of angle ACB which intersects the perpendicular bisector of BC at P.
vi) Join PB and draw

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Construct a triangle ABC, in which ABC=75o,AB=5 cm and BC=6.4 cm. Draw perpendicular bisector of side BC and also the bisector of Angle ACB.If these bisectors intersect each other at point P;prove that P is equidistant from B and C; and also from AC and BC.

Construct a triangle ABC, in which $A B C = 75^{o}$ ,AB=5 cm and BC=6.4 cm. Draw perpendicular bisector of side BC and also the bisector of Angle ACB.If these bisectors intersect each other at point P;prove that P is equidistant from B and C; and also from AC and BC.