In Triangle ABC; angle A = 60 degree ,angle C = 40 and bisector of angle ABC meets side AC - Maths -

Question

In Triangle ABC; angle A = 60 degree ,angle C = 40 and bisector
of angle ABC meets side AC at point P show that BP = CP

Ayush Jha · Accepted Answer

We have

Apply angle sum property of triangle we get $∠ A B C = 180 - (60 + 40) = 80 S i n c e B P i s a n g l e b i \sec t o r ∠ P B C = 40^{0} ∠ P C B = 40^{0} (g i v e n) S O, Δ P B C i s o c e l e s t r i a n g l e H e n c e, B P = C P$

In Triangle ABC; angle A = 60 degree ,angle C = 40 and bisector of angle ABC meets side AC at point P. show that BP = CP.