A line drawn through center of $\triangle ABC$ cuts sides $AB$ and $AC$ at $M$ and $N$, respectively. Prove $AM\cdot NC + AN\cdot MB = AM\cdot AN$

Question

I did Menelaus in $\triangle ACR;\triangle ABQ$, and I wrote $AM\cdot NC=2\cdot AN\cdot MR$ and $AN\cdot MB=2AM\cdot NQ$, so I added the equations and got $AM\cdot NC+AN\cdot MB=2(AM\cdot QN+AN\cdot MR)$. I cant see how to prove that 2(AM.QN+AN.MR)=AM.NA Please, can somebody help me? Thanks for antetion.

Answer 1

Let [.] denote areas. Then,

$$\frac{AM}{AB} = \frac{[AMG]}{[ABG]}= \frac{[AMG]}{\frac13[ABC]}=\frac{3[AMG]}{[ABC]},\>\>\>\>\>\>\>\frac{AN}{AC}=\frac{3[ANG]}{[ABC]}$$

Add the two equations and recognize $[AMG]+[ANG]=[AMN]$

$$ \frac{AM}{AB} + \frac{AN}{AC}=\frac{3[AMN]}{[ABC]}=\frac{3\frac{AM}{AB}\frac{AN}{AC}[ABC]}{[ABC]}=3\frac{AM}{AB}\frac{AN}{AC} $$

Rearrange to get

$$AM\cdot AC + AN\cdot AB = 3AM\cdot AN$$

Replace $AC = AN+NC$ and $AB = AM+MB$ to obtain $$AM\cdot NC + AN\cdot MB = AM\cdot AN$$