Interesting problem

[ins-]

Let $ABCD$ is a complete quadrilateral with intersection point of the diagonals -$P$. $E$ is the intersection point of $AB$ and $CD$. $F$ is the intersection point of $AD$ and $BC$. $X$ is a random point (outside triangle $AEF$). $A'$, $B'$, $C'$, $D'$, $P'$ are the intersection points of $AX$, $BX$, $CX$, $DX$, $PX$ with $EF$, respectively. Prove that:
a) $\frac{A'X}{AA'}+\frac{C'X}{CC'}=\frac{B'X}{BB'}+\frac{D'X}{DD'}$;
b) $\frac{A'X}{AA'}+\frac{B'X}{BB'}+\frac{C'X}{CC'}+\frac{D'X}{DD'}=4 \frac{P'X}{PP'}$.

(Оффтоп)

Greetings to Dimoniada :)

[DrVirogov]

ins- в сообщении #675610 писал(а):

...
$X$ is a random point (outside triangle $AEF$).
...

The restriction is not necessary. The point $X$ may be placed anywhere on the plain. Of course, relations may become negative, and it must be taken into consideration.

[ins-]

The problem I posted can be generalized in at least two more directions.

[DrVirogov]

No doubt it is so. The problem itself is generalisation of the sentence sometimes referred as "теорема о приставных лестницах ". There are multidimensional generalisations as well.

[ins-]

It will be interesting to see a multidimentional generalization.
What I had as ideas for generalizations were:
1. to construct a line parallel to EF
2. to construct lines through intersection point of the diagonals and to take their intersection points with the sides.

[ins-]

A method with name "равномерное движение" probably can be used to solve the problem, too. This problem was inspired by some works of Bulgarian mathematician Borislav Mihailov and one of my previous problems I hope you like it.

[DrVirogov]

Here is full 3D analog:
Let
ABCD be a simplex,
points $E,F,G$ lay on edges $AB,AC,AD $ respectively,
$Z$ be the point of intersection of the planes $BCG,CDE$ and $DBF$,
$P$ be the point of intersection $AZ$ and plane $EFG$,
$X$ be an arbitrary point (ignoring degenerate cases),
$A',E',F',G',Z',P'$ be the points of intersection $AX,EX,FX,GX,ZX,PX$ and plane $BCD$,
then
a) $\frac{E'X}{AE'}+\frac{F'X}{AF'}+\frac{G'X}{AG'}=\frac{Z'X}{AZ'}+2\frac{A'X}{AA'}$
b) $\frac{E'X}{AE'}+\frac{F'X}{AF'}+\frac{G'X}{AG'}+\frac{Z'X}{AZ'}+2\frac{A'X}{AA'}=6\frac{P'X}{PP'}$

[ins-]

What does "simplex" mean?

[DrVirogov]

In 3D space "simplex" means triangular pyramid.

[ins-]

Thank you for the nice generalizations.