Prove that the internal bisectors of the angles of a triangle are concurrent
Assertion (A) : AD is angle bisector of angleA of the triangle ABC. If AB = 6 cm, BC = 7 cm, AC = 8 cm then BD = 3 cm and CD = 4 cm. Reason (R) : The angle bisector AD of the triangle divides base BC in the ratio AB : AC.
Prove the following statement. "The bisector of an angle of a triangle divides the sides opposite to the angle in the ratio of the remaining sides"
The internal angle bisector of an angle of a triangle divide the opposite side internally in the ratio of the sides containgthe angle
In a triangle ABC, D is a point on BC such that AD is the internal bisector of angle A . Let angle B = 2 angle C and CD = AB. Then angle A is
If the bisector of the angle A of triangle ABC meets BC in D, prove that a=(b+c)[1-([AD]^(2))/(bc)]^((1)/(2))
If the bisector of angle A of the triangle ABC . makes an angle theta with BC, then sin theta=
In triangle ABC, AB = 12 cm, angle B = 60^@ , the perpendicular from A to BC meets it at D. The bisector of angle ABC meets AD at E. Then E divides AD in the ratio
In a Delta ABC, it is give ttha AD is the internal bisector of angleA . If AB=10 cm, AC=14 cm and BC=6 cm then CD=?
