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.
![geometry - In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF = geometry - In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF =](https://i.stack.imgur.com/B2F6m.png)
geometry - In $\Delta ABC$, $AB:AC = 4:3$ and $M$ is the midpoint of $BC$ . $E$ is a point on $AB$ and $F$ is a point on $AC$ such that $AE:AF =
![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 =
![The value of the determinant | b^2 - ab b - c bc - ac | ab - b^2 a - b b^2 - ab | bc - ac c - a ab - b^2 The value of the determinant | b^2 - ab b - c bc - ac | ab - b^2 a - b b^2 - ab | bc - ac c - a ab - b^2](https://haygot.s3.amazonaws.com/questions/2022827_1148955_ans_b6064ae55b644c44a4e9eb584afd2090.jpeg)