Find :$\frac{A1}{A}−\frac{A}{A2}$ in the triangle below
Solution 1:
Triangle $ABC$ and $MBN$ are similar, so $$\frac{AC}{MN}=\frac{a+b}a$$ Since $A1$ and $A$ have the same height, $$\frac{A1}{A}=\frac{a+b}a$$ If you draw a height from $B$ to $MN$, say $H$, and to $AC$, say $K$, then $$\frac{BH}{HK}=\frac ab$$ Note that the height from $A$ to $MN$ is equal to $HK$, so $$\frac A{A2}=\frac ba$$ Then $$\frac{A1}A-\frac A{A2}=\frac{a+b}a-\frac ba=\frac aa=1$$