Lemma 1

Quote:

Let $ABCD$ be a trapezoid $(AB \parallel CD)$ .
Let the points $E\in AD \cap BC$ and $I \in AC \cap BD$ .
$M,N$ are the midpoints of $AB,CD$ respectively.
It is known that the points $E,M,I,N$ are collinear.

Prove that the division $(E,M,I,N)$ is harmonic (that is, $\frac{EM}{EN}= \frac{IM}{IN}$ )

P.S. It can be applied also to a parallelogram $ABCD$

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Lemma 2

Quote:

Let $(E,M,I,N)$ be an harmonic division with ratios $\frac{EM}{EN}= \frac{IM}{IN}= \lambda$ .

$Z$ is a point $\in (MI)$ which divides the segments $MI,NE$ in the same ratio $\frac{ZM}{ZI}= \frac{ZN}{ZE}= \mu$

Then $\boxed{\mu = \frac{1}{2}(\frac{1}{\lambda}-\lambda)}$

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Proof
Suppose that $AB \parallel CD$ and $IC^{2}=IA \cdot AC$ . We will show that the triangles $EDC,IAB$ have the same centroid

from the Lemma 1 we get that $\frac{EM}{EN}= \frac{IM}{IN}= \lambda$ , for some $\lambda>0$

From the parallel lines $AB,CD$ we get that $\frac{IA}{IC}= \frac{IM}{IN}$

So, $IA = \lambda \cdot IC$

$\boxed{IC^{2}=IA \cdot AC}\Rightarrow$

$\lambda^{2}\cdot IC^{2}= \lambda \cdot IC(1+\lambda) IC \Rightarrow$

$\lambda^{2}= \lambda(1+\lambda) \Rightarrow \boxed{\lambda = \frac{\sqrt{5}-1}{2}}$ , because $\lambda>0$

As yetti noticed, $\lambda=\frac{1}{\phi}$ , where $\phi$ is the golden ratio.

From the Lemma 2 we get that the point $Z$ wich divides the segment $MI$ in the ratio $\mu=\frac{1-\lambda^{2}}{2\lambda}$ , divides the segment $NE$ in the same ratio

$\lambda=\frac{1}{\phi}\Rightarrow$
$\mu=\frac{1}{2}(\frac{1}{\lambda}-\lambda) = \frac{1}{2}(\phi-\frac{1}{\phi})=\frac{1}{2}$ .

So $\frac{ZM}{ZI}= \frac{ZN}{ZE}= \frac{1}{2}$ , which means that the point $Z$ is the centroid of $\triangle IAB$ and $\triangle ECD$ .

(=>)

We suppose that the triangles $EDC$ and $IAB$ have the same centroid.
We'll prove that $AB\parallel CD$ and $IC^{2}=IA \cdot AC$

Let $G$ be te common centroid.

From the Gauss line theorem we get that the points $M,N$ and the midpoint of $EI$ are collinear.

But $\frac{GM}{GI}= \frac{GN}{GE}\Rightarrow MN \parallel EI$

The midpoint of $EI$ belongs both to the lines $MN,EI$ , so the only possible result is that the points $M,N,E,I$ are collinear.

So the median $EN$ of the triangle $ECD$ is also median for the triangle $EAB$ .
And $A\in (ED, B\in (EC$
This concludes that $AB \parallel CD$ . So $ABCD$ is a trapezoid.

According to the Lemma 1 we have $\frac{EM}{EN}= \frac{IM}{IN}= \lambda$ , for aome $\lambda>0$
According to the Lemma 2, there is a unique point $Z$ in $(MI)$ which divides the segments $MI,NE$ at the same ratio, and the ratio is $\mu = \frac{1-\lambda^{2}}{2\lambda}$

$G$ is the one with this property, and the value of $\mu$ is $\frac{1}{2}$

$\frac{1}{2}(\frac{1}{\lambda}-\lambda) = \frac{1}{2}\Rightarrow$

$\frac{1}{\lambda}-\lambda) = 1 \Rightarrow$

$\lambda^{2}+\lambda-1 = 0 \Rightarrow \lambda = \frac{-1+\sqrt{5}}{2}$ , because $\lambda$ is positive

So $\frac{IA}{IC}= \frac{-1+\sqrt{5}}{2}$ , which yields $IC^{2}=IA \cdot AC$