Geometric proof

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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

 

[Deleted member 4993]

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

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[canvas]

AS*BC>=AC*BS+AB*SC
We know that BS=SC, so
AS*BC>=BS(AB+AC)=SC(AB+BC)
I'm stuck here

 

[Dr.Peterson]

AS*BC>=AC*BS+AB*SC
We know that BS=SC, so
AS*BC>=BS(AB+AC)=SC(AB+BC)
I'm stuck here

First, check your first line. Is that supposed to be Ptolemy's Theorem?

Then, comparing the corrected version of what you show here to the goal, we would like to show that AS*BC/SC < 2AS. Can you show that?

 

[canvas]

There should be BS(AB+AC)>=AS*BC
I don't know how to prove it, that's why I ask for it

 

[Dr.Peterson]

There should be BS(AB+AC)>=AS*BC
I don't know how to prove it, that's why I ask for it

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.

I hope you notice that AS*BC/SC <= 2AS is equivalent to BC/SC <= 2, and therefore BC <= 2SC.