5
$\begingroup$

The original problem can be found here: Nick's Mathematical Puzzle 62: Four squares on a quadrilateral : Squares are constructed externally on the sides of an arbitrary quadrilateral. Show that the line segments joining the centers of opposite squares lie on perpendicular lines and are of equal length.

$ABCD$ is an arbitrary quadrilateral; $E,F,G$ and $H$ are centers of squares outside the quadrilateral. Prove $EF\bot GH$ and $\overline{EF}= \overline{GH}$. enter image description here

The solution presented is using the geometric meaning of complex numbers.

Is there any pure geometric approach to prove it?

$\endgroup$
  • $\begingroup$ thanks! I have updated it with a brief comment. $\endgroup$ – LCFactorization Nov 9 '14 at 8:34
  • 1
    $\begingroup$ You surely mean $\overline{EF} = \overline{GH}$, not $\overrightarrow{EF} = \overrightarrow{GH}$. $\endgroup$ – user187373 Nov 9 '14 at 10:31
  • 2
    $\begingroup$ @LCFactorization I think you did not read the page entirely. There is a link on the page you cited labeled "a purely geometrical proof." $\endgroup$ – rschwieb Nov 10 '14 at 3:14
  • $\begingroup$ You're right. Thanks! They use different method to prove the lemma here. $\endgroup$ – LCFactorization Nov 10 '14 at 3:19
  • $\begingroup$ The overall context for this problem is "The Fundamental Theorem of Directly Similar Figures", e.g. zacharyabel.com/papers/Mean-Geo_talk_A05_McN_slides.pdf or zacharyabel.com/papers/Mean-Geo_A07.pdf $\endgroup$ – brainjam Jun 27 '18 at 18:26
3
$\begingroup$

Obviously, we can assume that $ABCD$ is oriented counterclockwise as in the figure. (Otherwise reflect through some line.)

For any point $M$ we write $\rho_M$ for the counterclockwise $90^{\circ}$ rotation about $M$, and $\sigma_M$ for the symmetry centred at $M$.

Lemma Given two points $M$ and $N$, we have $\rho_N \circ \rho_M = \sigma_O$ for some point $O$. Moreover $\rho_O(N) = M$.

Proof It is well-known that a composite of two rotations is a rotation whose angle is the sum of the oriented angles of the individual rotations (or a translation if this sum is $0$.) Thus there is a point $O$ for which $\rho_N \circ \rho_M = \sigma_O$.

Now write $M' = \sigma_O(M)$ and $N' = \sigma_O(N)$. $MNM'N'$ is a parallelogram with centre $O$. But since $$\rho_N(M) = \rho_N \circ \rho_M(M) = \sigma_O(M) = M',$$ the parallelogram $MNM'N'$ is in fact a clockwise-oriented square. The lemma follows immediately from this.

Proof of the theorem Let $O$ be the midpoint of the segment $AC$.

We have $\rho_H \circ \rho_E = \sigma_{O'}$ for some point $O'$ as in the lemma. But since $\sigma_{O'}(A) = \rho_H \circ \rho_E (A) = C$, the point $O'$ must in fact be $O$. Thus $\rho_O(H) = E$.

Considering $\rho_G \circ \rho_F$ and exchanging the roles of $A$ and $C$, we similarly prove $\rho_O(G) = F$.

Putting these facts together, we see that $\rho_O$ transforms the segment $HG$ into $EF$. This proves what we want.

$\endgroup$
  • $\begingroup$ your proof is very useful but much abstract. It uses "well-known" but difficult to illustrate lemma. I would prefer to prove a lemma first which I will post in an answer with a photo. $\endgroup$ – LCFactorization Nov 9 '14 at 11:35
  • 1
    $\begingroup$ You are talking about the fact that the composite of two rotations is a rotation or a translation. Every isometry that preserves orientation is either a rotation or a translation. So a composite of rotations must be one of these. The fact that you add the angles of rotation is clear by following the image of a single line under composition. $\endgroup$ – user187373 Nov 9 '14 at 11:52
  • $\begingroup$ Basically, answers within the scope of primary geometry are expected. Rotations for general cases can be examined via homogeneous coordinates in projective geometry arxiv.org/pdf/1307.0998v3.pdf; I will accept your answer if there is no other response acceptable. $\endgroup$ – LCFactorization Nov 9 '14 at 12:26
  • $\begingroup$ I have solved it. $\endgroup$ – LCFactorization Nov 10 '14 at 2:18
3
$\begingroup$

This was not intended to be an answer; I posted it as an answer so that I could attach a figure for illustration. But now it solves the problem.

I would prefer to prove a lemma that for any $\Delta ABC$ (in the figure below), $\Delta MEF$ obtained is an isosceles right-angled triangle, where $M$ is the mid-point of $AC$. $E$ and $F$ are centers of squares.

Update 1

Now it seems the lemma can be proved by adding auxiliary conductors like below:

enter image description here

Then $\Delta ABG\cong\Delta BPQ \cong \Delta CGB$, this can easily lead to: $\overline{AM}=\overline{BN}, \overline{BM}=\overline{PN}$ and therefore:

$\Delta ABM\cong \Delta BPN$; similarly $\Delta BCM\cong\Delta QBN$

Then $\angle MEN$ and $\angle MFN$ are both right angles; and $\Delta MEN$ and $\Delta MFN$ are isosceles; $ENFM$ is a square.

Update 2

By applying lemma above it is easy to conclude that: $\Delta EOF $ and $ \Delta HOG$ are congruent to each other and one can be obtained by rotating another around $O$ by 90°.

enter image description here

$\endgroup$
  • $\begingroup$ Its quite easy to prove its isosceles but right angled, hmm? $\endgroup$ – Sawarnik Nov 9 '14 at 16:15
  • $\begingroup$ Do you mean to prove $\Delta ABM \cong \Delta BPN$ and $\Delta BCM \cong \Delta QBN$ first? Is it possible not to use sine theorem? If possible, it will be easy to prove $\angle MEN$ and $\angle NFM$ are right angles, and then $ENFM$ is a square! $\endgroup$ – LCFactorization Nov 10 '14 at 0:27
  • 1
    $\begingroup$ No, I relied on the fact that triangles $ABQ$ and $PBC$ are congurent, and you can get $EF=FM$ easily. The right angle however eluded me, nice answer! :) +1 $\endgroup$ – Sawarnik Nov 10 '14 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.