Can we prove this geometry question without too much calculation?

Solution 1:

It looks to be related with a dissection of a dodecagon in three squares:

Let we assume that the square has a unit side, $\widehat{CAE}=\theta$ and $H$ is the projection of $D$ on $AE$.
If we prove $\theta=30^\circ$, the claim easily follows. We have $AE=\sqrt{2}$ and $$ AH = \cos\left(\frac{\pi}{4}-\theta\right) = \frac{\cos\theta+\sin\theta}{\sqrt{2}}, $$ so $AF=\sqrt{2}(\cos\theta+\sin\theta-1)$. By the cosine theorem $$ BF^2 = 1+AF^2-2 AF\cos\left(\theta+\frac{\pi}{4}\right) $$ and $$ \frac{BG}{GF}=\frac{[BAG]}{[GAF]}=\frac{\sin\frac{\pi}{4}}{AF\sin\theta}, $$ hence $$ BG^2 = \left(\frac{\sin\frac{\pi}{4}}{\sin\frac{\pi}{4}+AF\sin\theta}\right)^2 BF^2 = \frac{5-2 \cos\theta-2\cos(2\theta)-6\sin\theta+2\sin(2\theta)}{(2-\cos(2\theta)-2\sin(\theta)+\sin(2\theta))^2}$$ and by equating $BG^2$ and $CE^2=4(1-\cos\theta)$ $$ \sin\theta=\frac{1}{2} $$ follows (not easily - better to have the help of a CAS).

Solution 2:

This answer owes much to Jack D'Aurizio's accurate diagram, and his observation that $\widehat{CAE} = 30^\circ$.

I'll repeat his diagram here, for ease of reference when following the first part of the proof:

In order to determine the location of the point $E$, whose uniqueness will be shown later, suppose first for the sake of argument that the claimed proposition is true.

(If you're happy to accept a rabbit out of a hat, you can skip this part of the argument - it's not strictly part of the proof at all - and start from the construction of points $F$, $G$, and $E$ by method (3) below.)

Then, $CE$ is parallel and equal to $BG$, so $CEGB$ is a parallelogram. Hence, $GE$ is parallel and equal to $BC$. But $BC$ is parallel and equal to $AD$, therefore $GE$ is parallel and equal to $AD$.

Thus, $GEDA$ is a parallelogram. Therefore, $DE$ is parallel to $AD$; and therefore, $\widehat{ADE} = 135^\circ$.

Because $GF \parallel CE$, and $\triangle ACE$ is isosceles, so is $\triangle AGF$, therefore $AF = AG$.

Using again the fact that $GEDA$ is a parallelogram, we have $AG = DE$.

But we are given that $DF = DE$. Hence, $DF = DE = AG = AF$, i.e. $\triangle ADF$ is isosceles.

Let the two base angles of $\triangle ADF$ both be equal to $\phi$. Then the external angle $\widehat{DFE} = 2\phi$. Because $\triangle DFE$ is isosceles, $\widehat{DEF} = 2\phi$, therefore $\widehat{EDF} = 180^\circ - 4\phi$, therefore $\widehat{ADE} = 180^\circ - 3\phi$.

But we established earlier that $\widehat{ADE} = 135^\circ$; therefore $\phi = 15^\circ$.

Since $\widehat{DAC} = 45^\circ$, it follows that $\widehat{EAC} = \widehat{FAG} = 30^\circ$.

Because $\triangle AGF$ is isosceles, $\widehat{AGF} = 75^\circ$. But this angle is external to $\triangle ABG$, and we already know that $\widehat{GAB} = 45^\circ$, therefore $\widehat{ABG} = 30^\circ$.

Since $\widehat{FAB} = 75^\circ$, it follows from the angle sum of $\triangle ABF$ that $\widehat{AFB} = 75^\circ$, so $\triangle ABF$ is isosceles, and $FB = AB$.

This confirms what Jack D'Aurizio's diagram suggests, viz. that $F$ is the apex of an equilateral triangle erected on $BC$ as base.

We can construct $E$: (1) by producing $AF$ to a distance equal to $AC$; or alternatively, (2) by obtaining $G$ as the point of intersection of $AC$ with $BF$, and then completing the parallelogram $BCEG$; or yet again, (3) by constructing another equilateral triangle inside the square erected on side $CD$ to the right of the square $ABCD$ shown, and constructing $E$ inside that square, in the same way as $G$ inside $ABCD$.

Forget now that we were assuming that the claimed proposition is true, and start afresh!

We can't assume that construction methods (1), (2) and (3) are equivalent. I choose method (3).

Please excuse this shamefully low-tech figure, but I'm not up to speed with drawing figures digitally:

Construct $F$ as the apex of an equilateral triangle $\triangle BCF$ erected on $BC$ as base. Draw $AC$ intersecting $BF$ in $G$.

Produce $BC$ to the point $K$ where $CK = BC$. Construct an equilateral triangle $\triangle CKL$ on $CK$ as base. Draw $DK$ intersecting $CL$ in $E$.

(In fact, the three lines $DK$, $CL$, and $AF$ produced are concurrent, but we don't know that yet, whence the use of a dashed line in the diagram.)

Arguing in roughly the reverse order to before, we now know that:

$\widehat{FBC} = 60^\circ$, therefore $\widehat{ABF} = 30^\circ$; and $AB = BF$, therefore $\triangle ABF$ is isosceles, therefore $\widehat{FAB} = 75^\circ$; therefore $\widehat{FAG} = 30^\circ$ and $\widehat{DAF} = 15^\circ$.

Joining $FC$, we have a isosceles triangle $\triangle FCD$ congruent (by 'SAS') to $\triangle ABF$, so $DF = AF$.

Because $\widehat{FAG} = 30^\circ$ and $\widehat{AFG} = \widehat{AFB} = 75^\circ$, we have also $\widehat{AGF} = 75^\circ$, so $\triangle AGF$ is isosceles.

We use this fact in two ways:

First, we infer that $AG = AF$.

But we have already shown that $AF = DF$. Also, from the similarity of the methods of construction of points $E$ and $G$, the triangles $\triangle DCE$ and $\triangle ABG$ are congruent (the pattern is 'ASA': $DC = AB$, $\widehat{CDE} = 45^\circ = \widehat{ABG}$, and $\widehat{DCE} = 30^\circ = \widehat{ABG}$), therefore $DE = AG$.

Putting these three equations together, we get $DE = AG = AF = DF$.

In the isosceles triangle $\triangle DFE$, the apex angle $$\widehat{EDF} = \widehat{ADC} + \widehat{CDE} - \widehat{ADF} = 90^\circ + 45^\circ -15^\circ = 120^\circ $$ (alternatively, use the known fact that $\widehat{FDC} = 75^\circ$), therefore $\widehat{DFE} = 30^\circ$. Since $30^\circ$ is the sum of the two interior opposite angles $\widehat{DAF} = \widehat{FDA} = 15^\circ$ in isosceles triangle $\triangle DAF$, it follows that the point $E$ lies on the line $AF$ produced (shown dashed in the diagram).

We are now ready to make the second deduction from the fact that $\triangle AGF$ is isosceles:

From the construction of the points $E$, $F$, and $G$ using equilateral triangles standing on the line $BC$ produced (to $K$), because the lines $CE$ and $GF$ are parallel (they both make angle $60^\circ$ with the line $BCK$), the triangles $\triangle ACE$ and $\triangle AGF$ are similar. Therefore $\triangle ACE$ is isosceles, so $AE = AC$.

Therefore, with this particular construction of the points $E$, $F$, and $G$, all four of the conditions of the problem are satisfied. So is its conclusion, because, as has already been remarked (in the course of proving that the triangles $\triangle ACE$ and $\triangle AGF$ are similar), $BG \parallel CE$.

I was going to complete the argument by arguing that if the circle with centre $A$ and radius $AC$ intersects the line $AD$ produced at the point $M$, and if a point $P$ is taken anywhere on the arc $MC$, and then a point $Q$ constructed on the radius $AP$ with $DQ = DP$, and finally a point $R$ constructed as the intersection of $BQ$ with $AP$, the length $BR$ can be shown to increase strictly as $P$ moves from $M$ to $C$ (a loose form of words which can be dispensed with), while the length $CP$ decreases strictly, so the point $P = E$ is the only one where $BR = CP$. It follows that there is only one triple of points ($E$, $F$, $G$) satisfying the conditions of the problem; and then also $BG \parallel CE$.

But I see that dxiv has now posted an answer which ends with the same argument. I'd still like to spell this part of the proof out in more detail, but it can wait, and this answer is already long enough!

Solution 3:

The following will (1) solve a closely related problem, then (2) show that it's equivalent to the one originally posed.

(1) Using the same picture as posted but with a different construction, let $ABCD$ be the square, and $G$ a point on the diagonal $AC$. Define point $E$ such as $GE$ is parallel to $BC$ and $GE=BC$ (so that both $BCEG$ and $AGED$ are parallelograms). Let $F$ be the intersection of $BG$ and $AE$.

It will be proved that: there exists a unique point G for which $AE = AC$, and for that point (only) it also holds true that $DF = DE$.

Calculations work out relatively easy in a complex plane set with the origin at A, and unit $AD$ along the real axis. Then (using $z$ for the complex number associated with point $Z$):

$$a=0, \;\;b=-i, \;\;c = 1-i, \;\;d=1, \;\;g = \lambda c \;\text{ with }\lambda \in (0,1) $$

Since $AGED$ is a parallelogram: $e = d + (g - a) = (1 + \lambda) - i \lambda$. Then:

$$ \;\;AE^2 - AC^2 = |e-a|^2 - |c-a|^2 = (1 + \lambda)^2 + \lambda^2 - 2 = 2 \lambda^2 + 2 \lambda - 1 $$

$$\begin{align} \;\;DF^2 - DE^2 = |f-d|^2 - |e-d|^2 & = |(\lambda^2 + \lambda - 1) - i \lambda^2|^2 - |\lambda - i \lambda|^2 \\ & = 2 \lambda^4 + 2 \lambda^3 - 3 \lambda^2 - 2 \lambda + 1 \\ & = (\lambda^2 - 1)(2 \lambda^2 + 2 \lambda - 1) \\ & = (\lambda^2 - 1)(AE^2 - AC^2) \end{align} $$

$\lambda^2 - 1 \ne 0$ since $\lambda \in (0,1)$ , which proves that $\;\;DF = DE \;\; \iff \;\; AE = AC$.

Furthermore, since $2 \lambda^2 + 2 \lambda - 1$ is strictly increasing on $(0,1)$ the point $G$ for which $AE=AC$ and $DF=DE$ is uniquely determined (in fact, it's the $G$ corresponding to $\lambda = (-1 + \sqrt{3})/2$).

(2) Back to the original problem, ignore for a moment the last $BG=CE$ constraint, and let $A'$ be the point on line $AD$ such as $AA' = AC$ on the same side of $A$ as $D$. It is easy to check that $BG \lt CE$ at $E \equiv A'$, and $BG \gt CE = 0$ at $E \equiv C$, so there must exist at least one point $E$ along the way where $BG=CE$. It can also be shown that, as point $E$ moves from $A'$ to $C$ along the arc of the circle $AE = AC$, $BG$ increases and $CE$ decreases, so the point where $BG=CE$ is unique. Finally, that unique point $G$ matches the same constraints as the unique point $G$ determined at (1) so in the end they must be the same, thus $BG \parallel CE$ by the construction at (1).