In triangle ABC , AB =AC and BC=AB +AI , where I is the incentre of triangle ABC . Then find the measure of angle A.
In ∆ABC, if AC is greater than AB, then prove that AC AB is less than BC, AC BC is less than AB and BC AB is less than AC.
![SOLVED: Expression a + 0 =a a+b=b+a Dual a . 1 =0 ab ba a+(b+c)=latb)tc a(bc) = (abc a + bc = Ka+ ba +c) a(b +c) = ab + ac a+6 = SOLVED: Expression a + 0 =a a+b=b+a Dual a . 1 =0 ab ba a+(b+c)=latb)tc a(bc) = (abc a + bc = Ka+ ba +c) a(b +c) = ab + ac a+6 =](https://cdn.numerade.com/ask_images/27d2cf9107834440955a2edaf77aafc2.jpg)
SOLVED: Expression a + 0 =a a+b=b+a Dual a . 1 =0 ab ba a+(b+c)=latb)tc a(bc) = (abc a + bc = Ka+ ba +c) a(b +c) = ab + ac a+6 =
![SOLVED: P and Q are the points on the side BC of triangle ABC and AP=AQ prove that :AC+AB+BC is greater than 2AP +PQ SOLVED: P and Q are the points on the side BC of triangle ABC and AP=AQ prove that :AC+AB+BC is greater than 2AP +PQ](https://cdn.numerade.com/ask_previews/85447aea-613e-4130-85b6-88231d8d83ff_large.jpg)