I have a vector proof of the following problem, but the synthetic proof seems elusive:
$ABCD$ is a convex quadrilateral. The line $AB$ intersects the line $CD$ at $P$. If $M,N$ are mid points of segments $BD$ and $AC$ respectively. Then prove that $4[PMN] = [ABCD]$.
Notation: [PQRS] denotes the area of the polygon PQRS.
I have been juggling around with marking midpoints and moving line segments parallel to other line segments. I just can't find a synthetic proof.
We assume that the order $A,B,C,D$ is taken in the direct (anticlockwise) orientation.
We are going to use property of determinants such as: a) multiplication of a column by $\lambda$ amounts to multiply its value by $\lambda$, b) interchanging two columns yields a sign change, c) $\det(\vec{u}+\vec{v},\vec{w})=\det(\vec{u},\vec{w})+\det(\vec{v},\vec{w})$, etc.
A preliminary recall: for any triple of points $M,N,P$:
$$\det(\vec{MN},\vec{MP})=2 [MNP] \ \ \text{with the meaning of oriented area}.$$
(i.e., + sign is $MNP$ has direct orientation, -1 otherwise).
Let us express $\det(2 \vec{PM},2 \vec{PN})$ in two different ways:
$$\tag{1}\text{1st way:} \ \ \det(2 \vec{PM},2 \vec{PN})=4 \det(\vec{PM},\vec{PN})=8[PMN]$$
$$\text{2nd way:} \ \ \det(2 \vec{PM},2 \vec{PN})=$$
$$=\det(\vec{PB}+\vec{PD},\vec{PA}+\vec{PC})=$$
$$=\underbrace{\det(\vec{PB},\vec{PA})}_{0}+\det(\vec{PB},\vec{PC})+\det(\vec{PD},\vec{PA})+\underbrace{\det(\vec{PD},\vec{PC})}_{0}=$$
$$\tag{2}=\det(\vec{PB},\vec{PC})-\det(\vec{PA},\vec{PD})=2[PBC]-2[ADP]=2[ABCD].$$
Equating (1) and (2) gives the result.
Edit; there is a second method which is more in the spirit of a synthetic method. One may be surprised to see coordinates. But there is no contradiction. As we are referring to methods that Greeks would have used, we are precisely with these methods: it is not well known that, although Greek mathematics do not use "true" coordinates, they are rather close to it in many works, and using two coordinates $(a,b)$ for a point with respect to a basis-parallelogram, is like saying that the areas' ratio of 2 parallelograms is $ab$.
Consider the previous figure with the following coordinates:
The area of triangle $PDA$, resp. $PCB$, is $da/2$ resp. $bd/2$.
Thus the area of quadrilateral $ABCD$ is $(ad-bc)/2$.
Besides, the area of triangle $PMB$ is the half of the following determinant: $\det \pmatrix{d/2&c/2\\b/2&a/2}$, i.e., $\frac18 (ad-bc)$ which is in the ratio 1:4 with the previous result, as desired.
Remark 1: I don't imagine a quicker proof...
Remark 2: One may object that the concept of area has not its place in affine geometry, but in fact, what we are looking for is a ratio of areas, and this is an affine concept (for example barycentric coordinates can be represented as ratios of areas).