Draw two lines g and h. Then draw three points A, B, C on line g and 3 points a, b, c on line h. Let X be the intersection of lines Ab and Ba, Y of Ac and Ca, and Z of Bc and Cb. Pappus's theorem says points X, Y and Z are collinear. Can you see why?

technically, the below can be a proof why the theorem is true, but I know that it's the lowest form of proof. so, I still don't know how to prove the theorem in an intuitive way. However, let's try this anyway:

without loss of generality, we can put one of line g and h as x axis then we can say the equation of the line is x=0, and let's put the equation of the other line as y=px. let's put six points like this: (a, 0), (b, 0), (c, 0), (A, pA), (B, pB), (C, pC), and put the intersection point X as (X1, Y1), Y as (X2, Y2), and Z as (X3, Y3)

first, let's find X. you can put two lines like the below:

y = (x-a)pB/(B-a)
y = (x-b)pA/(A-b)

then we can get X1 and Y1 (omit details):

X1 = (AB(a-b)+ab(A-B))/(Aa-Bb)
Y1 = pAB(a-b)/(Aa-Bb)

the other intersection points can be gotten by replacing characters:

X2 = (AC(a-c)+ac(A-C))/(Aa-Cc)
Y2 = pAC(a-c)/(Aa-Cc)
Y3 = pBC(b-c)/(Bb-Cc)
X3 = (BC(b-c)+bc(B-C))/(Bb-Cc)

what we want to show is (X1-X2, Y1-Y2)//(X1-X3, Y1-Y3) which can be shown with the determinant:

| X1 - X2 Y1 - Y2 |
| | = 0
| X1 - X3 Y1 - Y3 |

the calculation of the above is really tiresome, so we can use a computer to do it:

$ python << EOF
from sympy import *
p, a, b, c, A, B, C = symbols("p, a, b, c, A, B, C")
X1, X2, X3, Y1, Y2, Y3 = symbols("X1, X2, X3, Y1, Y2, Y3")
X1 = (A*B*(a-b)+a*b*(A-B))/(A*a-B*b)
X2 = (A*C*(a-c)+a*c*(A-C))/(A*a-C*c)
X3 = (B*C*(b-c)+b*c*(B-C))/(B*b-C*c)
Y1 = p*A*B*(a-b)/(A*a-B*b)
Y2 = p*A*C*(a-c)/(A*a-C*c)
Y3 = p*B*C*(b-c)/(B*b-C*c)
print("the result is ", simplify((X1-X2)*(Y1-Y3) - (Y1-Y2)*(X1-X3)))
EOF

the result is 0, so the theorem holds.

do you have any hint to prove this in an intuitive way? I don't think Pappus went through like the above way... he probably had a better and ingenious idea. please enlighten me.

P1527
I tried it myself and what I've come up with so far is pretty similar. The main thing I did differently was to do a few more transformations to make things as simple as possible before starting on the algebra.

This theorem holds in a plane if and only if it holds in a perspective drawing of a plane. A perspective drawing transforms parallel lines to lines that meet at a vanishing point. We can reinterpret the picture as a perspective drawing of a plane in which lines g and h are parallel, and so are lines Aa and Cc.

Linear transformations also don't affect whether the theorem holds, so we can stretch and skew the parallelogram to make a square with the points at:
(-1, 1) (1, 1)
(-1, a) (1, b)
(-1, -1) (1, -1)

Then we can do the algebra so show the intersection points are at
( (b-a)/(2-a-b), (1-ab)/(2-a-b) )
( 0, 0 )
( (a-b)/(a+b+2), (ba-1)/(a+b+2) )
which are clearly collinear.

In addition to the coordinate style proofs like ours, Wikipedia has some stuff on what Pappus did:
https://en.wikipedia.org/wiki/Pappus%27s_hexagon_theorem#Origins
And apparently Pappus's Collection is the place for further reading.

The third video in P1442 talks about yet another way to prove it with techniques from geometric algebra, which would certainly not be what Pappus used, but still worth looking at.

I was wondering if there was a tricky method like if we imagine X, Y, Z in C, then Y = X + t(Y-X) at t=1 tautologically, and all the intersection rules imply exists real t such that Z = X + t(Y-X). I guess in polar form with the original at X, then Y and Z will turn out to have the same angle. I'm just describing doing the work other people did above though.

P1541
X is 0+0j. Without loss of generality we can put Y at 1+0j.
A=d cis e, R contains d>0, 0<=e<2pi
b=f cis e, R contains f<0
B=g cis h, R contains g>0, 0<=h<2pi
a=k cis h, R contains h<0
And then the problem actually begins, I will try it later.

Problems like this bring me back to high school math olympiads :)

The elementary proof I know of was using menelaus' theorem, see here: https://www.cut-the-knot.org/pythagoras/Pappus.shtml

There's a similar proof of its generalization, pascal's theorem, also using menelaus, also on the same website.

I swear to god I remember learning some kind of super elegant proof of this from Dušan Djukić (Yugoslavian IMO coach I used to know). I think it used some kind of circle theorem, like Miquel point or radical axes or something. But with no cyclic quads in sight...it could very well have just been the menelaus proof.

Anyway, >>P1540 is probably the better way to go about it, I just like elementary solutions to geometry puzzles for sentimental reasons.

P1548
I don't remember your fancy math names, but a circle idea would be to show that the points where the circles described by
AYC aYc XBZ XbZ
touch form a line.

P1549
To stay on theme you could show that they cannot possibly lie on the perimeter of the same circle ;p

P1548
That's pretty cool. Personally I found Pappus' original proof using cross ratios a little easier to grasp the essence of. But the concepts seem pretty similar; I wonder if there's some relation between cross ratios and Menelaus' theorem besides both being useful to prove Pappus.

This page about applying Pappus' theorem to games looks very interesting and it's a shame the Java applet doesn't work in browsers anymore. Maybe I'll have to get it working outside the browser and play with it.
https://www.cut-the-knot.org/Curriculum/Games/MixedStrategies.shtml