# Parallel lines in parallelogram inside parallelogram structure

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Show that GF // HC When I tried to prove another geometry question, it brought me the following figure. In the figure, ABCD is a parallelogram. E is a point on AD and F is a point on BC such that EB // DF. G is a point on AB. DG intersects EB at H. Show that GF // HC.

I found geometrical solution to this problem and will give my answer below

We can work with square ABCD instead of parallelogram (and BFDE will still be a parallelogram) pointing out that square can be transformed to any parallelogram by suitable affine transform that will not affect parallelism of lines and will not change ratios in which points E and G divide AD and AB. With square many things look simpler. I also rotated picture by 90 degrees because I want to show that lines HC and GF undergo shear transformation parallel to line BC when points G and H move along AB and EB and it’s easier to present it that way. We know that if G coincides with point I on my picture, which is symmetric to point E with respect to diagonal AC of square ABCD, lines HC and GF become parallel as they coincide with lines JC and IF. If we can show that with any other position of G on side AB, lines HC and GF can be presented as images of lines JC and IF under shear transformation parallel to line BC, this will mean that lines HC and GF are always parallel. There are two cases to be considered – first $$case A$$ is when G belongs to IB and second $$case B$$ is when G belons AI. Let’s consider fist case, I denote distances of points from line BC as $$1$$ for A, $$h$$ for J, $$2*h – 1$$ for P, $$q$$ for H, $$s$$ for G and $$0$$ for B, also denote AE as $$r$$. Among these only $$h$$ and $$q$$ are inputs of the problem, $$h$$ fixes position of E on AD (and F on BC) and $$q$$ fixes position of H on BE and G on AB, all others can be calculated as shown below based on triangle similarity. $$r = (1 - h) / h$$ triangles AEJ and KHJ: $$KH = r * (h - q) / (1 - h) = (h - q) / h$$ triangles BHN and BEA: $$HN = q * r = q * (1 - h) / h$$ triangles GHN and GDA: $$s = (q - HN) / (1 - HN) = (q - q * (1 - h) / h) / (1 - q * (1 - h) / h) = ( q * h - q * (1 - h) ) / ( h - q * (1 - h) ) = (2 * q * h - q ) / ( h - q + q * h )$$ triangles MLP and HOP: $$LM = (KH - r) * (2 * h - 1 - s) / (2 * h - 1 - q) = ( (h - q) / h - (1 - h) / h ) * (2 * h - 1 - s) / (2 * h - 1 - q) = ( (2 * h - q - 1 ) / h ) * (2 * h - 1 - s) / (2 * h - 1 - q) = (2 * h - 1 - (2 * q * h - q ) / ( h - q + q * h ) ) / h = (2 * h - 1) * (1 - q / ( h - q + q * h ) ) / h = (2 * h - 1) * ( h - q + q * h - q ) / h / ( h - q + q * h ) = (2 * h - 1) * ( h - 2 * q + q * h ) / h / ( h - q + q * h )$$ triangles BHN and BMG: $$MG = HN * s / q = ( q * (1 - h) / h ) * s / q = s * (1 - h) / h = ( (2 * q * h - q ) / ( h - q + q * h ) ) * (1 - h) / h = ( 2 * q * h - q ) * (1 - h) / h / ( h - q + q * h )$$ $$LG = MG + LM = ( 2 * q * h - q ) * (1 - h) / h / ( h - q + q * h ) + (2 * h - 1) * ( h - 2 * q + q * h ) / h / ( h - q + q * h ) = ( ( 2 * q * h - q ) * (1 - h) + (2 * h - 1) * ( h - 2 * q + q * h ) ) / h / ( h - q + q * h ) = ( q * (1 - h) + ( h - 2 * q + q * h ) ) * (2 * h - 1) / h / ( h - q + q * h ) = ( q - q * h + h - 2 * q + q * h ) * (2 * h - 1) / h / ( h - q + q * h ) = ( h - q ) * (2 * h - 1) / h / ( h - q + q * h )$$ $$LG / KH = ( h - q ) * (2 * h - 1) * h / h / ( h - q + q * h ) / (h - q) = (2 * h - 1) / ( h - q + q * h ) = s/q$$ This proves case A.

Case B $$r = (1 - h) / h$$ triangles AEJ and OHJ: $$OH = r * (q - h ) / (1 - h) = (q - h) / h$$ triangles BHN and BEA: $$HN = q * r = q * (1 - h) / h$$ triangles GHN and GDA: $$s = (q - HN ) / (1 - HN ) = (q - q * (1 - h) / h ) / (1 - q * (1 - h) / h ) = (q * h - q * (1 - h) ) / (h - q * (1 - h) ) = (2 * q * h - q ) / (h - q + q * h )$$ triangle GLI: $$GL = s - 1 + r = (2 * q * h - q ) / (h - q + q * h ) - 1 + (1 - h) / h = (2 * q * h * h - q * h - h * h + q * h - q * h * h + h - q + q * h - h * h + q * h - q * h * h ) / (h - q + q * h ) / h = ( - 2 * h * h + 2 * q * h + h - q ) / (h - q + q * h ) / h = (q - h ) * ( 2 * h - 1 ) / (h - q + q * h ) / h$$ $$GL / OH = (2 * h - 1 ) / ( h - q + q * h ) = s / q$$ This proves case B.

• I haven't read everything, but it's an impressive piece of work, to say the least. Commented Apr 11 at 17:30
• Would not do it again! I mean writing these formulas)) but reputation is worth doing it… Thanks! Commented Apr 11 at 17:41
• "A little" too complicated IMHO. Commented Apr 11 at 19:22
• @Jean Marie, sorry for being a bit philosophical, but I want to point on strong sides of my old style approach, firstly, if we compare mathematics to battlefield and mathematical problems to enemy, it makes no difference how you kill (solve) it, with old rusty axe of triangular similarities or with Swiss army knife of complex numbers, secondly, if we compare mathematics to sport, it may look “not sportive” to use complex numbers where stone axe is enough, and lastly, I could explain my proof to Pythagoras in ten minutes but if I would show him complex numbers and matrices he would curse me!) Commented Apr 12 at 12:41
• @Jean Marie, nice to hear that! my last point (on Pythagoras) was probably not well formulated, I do not mean that advanced mathematical tools should not be used and must be cursed, but I see advantage in using simple tools when possible because it is actually easier to explain long but simple proof to people with not much background in math (like me)), while formal simplicity of a proof that uses advanced tool is more than compensated by complexity of education one needs to have to follow such proof Commented Apr 12 at 15:42

We can indeed work with the square $$ABCD$$ because the square can be transformed with transvection and dilation into a parallelogram.

We then deliberately place ourselves in $$\mathbb C$$, with $$D=0,C=1,A=i,B=1+i, E=bi, G=a+i$$, where $$a,b\in \mathbb R$$

Then $$F=1+(1-b)i$$

Let us note $$DF=\boxed{l_0=\mathbb R[1+(1-b)i]}$$ $$BE=\boxed{l=1+i+l_0=1+i+\mathbb R[1+(1-b)i]}$$

$$DG=\boxed{m=\mathbb R(a+i)}$$

$$H\in l\cap m$$ is like solving a system with two unknowns $$\xi$$ and $$\eta$$ $$1+i+\xi[1+(1-b)i]=\eta(a+i)$$ $$\iff \begin{cases} \xi-\eta a=-1 \\ \xi(1-b)-\eta=-1 \end{cases}$$ $$\implies \eta=\frac{b}{1+ab-a}$$

So, $$\boxed{H-C=\frac{1}{1+ab-a}\color{red}{(a-1,b)}}$$

To conclude, It suffices to observe that $$\boxed{G-F=\color{red}{(a-1,b)}}.\square$$ https://www.geogebra.org/classic/wufg4swx

Note:

• We introduce $$A:=\begin{bmatrix}1 & -a \\1-b & -1\end{bmatrix}, \det A=-(1+ab-a)$$ ( to check these somewhat tedious calculations).

• Effect of a transvection on a square

• Good explanations. Commented Apr 11 at 19:25
• @JeanMarie : Thank you very much. I appreciate it very much, especially from you. Commented Apr 11 at 19:27
• @Stéphane Jaouen, Thanks for you answer! it looks much more concise then mine, I must admit, as I lack good working knowledge of complex numbers, can I ask you to look at this problem that I posted some days back?, do you think it can be solved the same way you did here? it was already deleted and if you want to see it but you can’t (i am new here)), let me know I will add it to my answer… math.stackexchange.com/questions/4889078/… Commented Apr 11 at 19:54
• @Stéphane Jaouen, understood! I have no idea what it “wrong” here… will learn it slowly, thanks for advice, sooner or later I will post it again if will not solve it myself Commented Apr 11 at 20:09