Trapezoids in a square

Good day

As part of a problem I need to show that AB is parallel to CD, with the given info on the image. All the segments marked red are equal, all 1-stripe grey equal etc.

I'd like to prove EA=AD, then it should prove AB||DC

Suggestions/tips? I'd appreciate pointers rather than a complete answer so as to try it self first!

Can we do the following?

We the right-angles on AC. Now we draw in AI parallel to EF. We then have parallelogram AIFH - thus $\angle AHF$ and $\angle AIF$ are both 90 degrees. Thus $\angle HAI$ = $\angle IFH$ = 90 degrees. Thus AB||DC ?

Place the square on a Cartesian coordinate grid. We can choose the units so the square is a unit square. The coordinates of the vertices $B,F,D,E$ are then obvious. (I limited this diagram to only what is necessary for my solution: your diagram has unneeded line segments and not enough labels for the points.)

Point $C$, the midpoint of $\overline{BF}$, has coordinates $(0.5,1)$. The line $\overleftrightarrow{CD}$ then has the equation $y=2x-1$.

According to your diagram, the point I have labeled $G$ in my diagram is a unit distance from point $B$, which is the origin. Point $G$ therefore lies on the unit circle $x^2+y^2=1$. We can solve those two simultaneous equations for line $\overleftrightarrow{CD}$ and the unit circle to get the only intersection point in the first quadrant (inside the square), and we get $G=(0.8,0.6)$.

The point I have labeled $H$, the midpoint of $\overline{EG}$, then has coordinates $(0.4,0.8)$. The line $\overleftrightarrow{BH}$ then has the equation $y=2x$.

Point $A$ is defined as the intersection of line $\overleftrightarrow{BH}$ with the top of the square, $y=1$, so the coordinates of $A$ are $(0.5,1)$.

We then easily see that $\overline{AB}\parallel\overline{DC}$: for one thing, the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{DC}$ have the same slope, namely $2$.

Now you can translate that argument into a purely Euclidean geometric argument, if you like.

• +1, great answer, could you please tell me what software you used to create the graphic? Does it by chance allow to define constraints of the elements of the graphic such as points, lines, etc?
– null
Commented Aug 24, 2015 at 22:50
• @null: I used Geogebra, which is right now advertised in the right column of this web page. it is very possible and does allow constraints on various objects. I defined the lines, circle, midpoints, intersection points, etc. within Geogebra, and it calculated the coordinates.I highly recommend it for work with geometry or algebra (the source of its name). Commented Aug 25, 2015 at 0:23
• Thank you for the contribution! You've sparked the idea for this kind of approach, I think I'll try it! Commented Aug 25, 2015 at 15:01

Borrowing @Rory Daulton's notation, construct a second square directly beneath the original, and extend the line $\overleftrightarrow{DC}$ to meet the lower square at vertex $K$.

The circle centered at $B$ with radius $BE$ passes through points $E$, $G$ and $K$. Therefore the inscribed angle $\angle EKG$ is half the central angle $\angle EBG$, since both angles subtend the same arc $\stackrel\frown{EG}$. But $\overline {BA}$ bisects the central angle $\angle EBG$, so angle $\angle EBA$ equals $\angle EKG$. This implies that $\overline {AB}$ is parallel to $\overline{KG}$, which coincides with $\overline {CD}$.

• This is a good answer, more geometric than mine. +1! Commented Aug 25, 2015 at 0:23