Mathematics
Bisector of angle B of triangle ABC intersects side AC at point P, then point P is :
equidistant from vertices A and C
PA = PB
PB = PC
equidistant from sides AB and BC
Related Questions
Given: CP is the bisector of angle C of ∆ABC.
Prove: P is equidistant from AC and BC.
Given: PQ is a perpendicular bisector of side AB of the triangle ABC.
Prove: Q is equidistant from A and B.
The perpendicular bisector of side AB and bisector of angle A of △ABC meet at point P. Then :
PA = PB
PA = PC
PB = PC
PB bisects ∠ABC
Using the information in the given diagram, state if :
AD = DC
BD = DC
CD bisects angle ACB
angle CAD is greater than angle DAB