Please show us what you have tried and exactly where you are stuck.Prove using ptolemy's theorem.
Given triangle ABC is circumscribed such that bisector of the angle BAC insersects the circle in point S, prove that AB+AC<2AS
First, check your first line. Is that supposed to be Ptolemy's Theorem?AS*BC>=AC*BS+AB*SC
We know that BS=SC, so
AS*BC>=BS(AB+AC)=SC(AB+BC)
I'm stuck here
Please draw a picture of the problem, and show us what you have. It is possible to see this part of the proof just by looking at the picture.There should be BS(AB+AC)>=AS*BC
I don't know how to prove it, that's why I ask for it