Find :$\frac{A1}{A}−\frac{A}{A2}$ in the triangle below

Question

For refrence: In the figure if $MN \parallel AC$ calculate:$\frac{A1}{A}-\frac{A}{A2}$(Answer:$1$)

My progress: $\frac{A}{A2} = \frac{AM}{BM}\\ \frac{A1}{A} = \frac{b}{AM}\\ \frac{A2+A}{a} = \frac{A1}{b}\\$

,,,I'm missing some relationship that I haven't seen yet.

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Answer 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 $$\frac{A1}A-\frac A{A2}=\frac{a+b}a-\frac ba=\frac aa=1$$