Prove that $\frac {AN}{ND}=\frac{1}{2} \cdot \frac{AM}{MD}$ elementary.

Question

Let $\triangle ABC$, $D\in BC, M\in AD$, $BM\cap AC=\{E\}, CM\cap AB=\{F\}, EF\cap AD=\{N\}$. Prove that $$\dfrac {AN}{ND}=\dfrac{1}{2} \cdot \dfrac{AM}{MD}$$

Now I have a projective solution and I would like to get nice elementary solution.

Projective solution: Let $EF$ meet $BC$ at $X$ and let it meet $AB$ and $AC$ at $G$ and $H$ respectively. Then

\begin{align} (A,D;M,N)&= (XA,XD;XM,XN)\\ &= (XA,XB;XG,XF)\\ &= (A,B;G,F)\\ &= (MA,MB;MG,MF)\\ &= (MA,ME;MH,MC)\\ &= (A,E;H,C)\\ &= (XA,XE;XH,XC)\\ &= (XA,XN;XM,XD)\\ &= (A,N;M,D) \end{align}

So we have $${AM\over MD}:{AN \over ND} = {AM\over MN}:{AD \over DN}$$ and thus $$AN\cdot MD = AD\cdot MN \implies xz=y(x+y+z)$$ which is equivalent to formula we want.

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Interesting note: If we move $M$ on $AD$ point $X$ is fixed:

By Menelaus theorem for triangle $ABC$ and transversal $X-F-E$ we have $${BX\over XC}\cdot {AF\over FB}\cdot {CE\over AE} =1\;\;\;/\cdot{CD\over DB}$$ so we have by Ceva theorem for $ABC$ and $M$: $${BX\over XC}\cdot \underbrace{{AF\over FB}\cdot {CE\over AE}\cdot{CD\over DB}}_{=1} ={CD\over DB}$$ So we see that as we move $M$ (and $N,E,F$) point $X$ is fixed.

Answer 1

Comment: some observations in figure: The figure is constructed such that $AM=MD$ and $AD\bot BC$, $BE\bot AC$ and $CF\bot AB$.

$\triangle MEF\sim \triangle MBC$

$\angle AFE=\angle ACB\Rightarrow \triangle AEF\sim \triangle ABC$

But it does not look that $\frac{AN}{ND}=\frac 12 \cdot \frac {AM}{MD}$

Using similarities we may find $\frac{AN}{ND}=\frac 1 k \cdot \frac {AM}{MD}$

The figure shows $k\neq 2$

Answer 2

The problem is purely projective in nature, so a solution may need to use projective arguments. I am trying to reduce them to a minimum, using only Menelaus and a simple property of the cross ratio, namely the relation between $(S,T,U,V)$ and $(S,U,T,V)$ on a line, which may be shown using cartesian coordinates on the line.

We apply Menelaus in:

• triangle $\Delta ADC$ with "secant" line $XNE$, and
• triangle $\Delta ADC$ with "secant" line $BME$:

$$ \begin{aligned} 1=\frac 11&= \frac {\displaystyle \frac{NA}{ND}\cdot \frac{XD}{XC}\cdot \color{gray}{\frac{EC}{EA}} } {\displaystyle \frac{MA}{MD}\cdot \frac{BD}{BC}\cdot \color{gray}{\frac{EC}{EA}} } = \frac {\displaystyle \frac{NA}{ND}:\frac{MA}{MD} } {\displaystyle \frac{BD}{BC}:\frac{XD}{XC} } \ , \\[3mm] &\qquad\qquad\qquad\qquad\text{ and the denominator in the above fraction is:} \\ \frac{BD}{BC}:\frac{XD}{XC} &=(B,X,D,C) =\frac 1{(X,B,D,C)} =\frac 1{1-(X,D,B,C)} \\ &=\frac 1{1-(-1)}=\frac 12\ . \end{aligned} $$ So the numerator in the same fraction is also $\frac 12$.

Here, we have used the fact that the $4$-tuple $(X,D;B,C)$ is harmonic. This can be shown by using Ceva and Menelaus for $\Delta ABC$ (for the cevians through $M$, and respectively for the "secant" line $XEF$.

$\square$