Equilateral Triangles On Sides of a Parallelogram

Question

On the sides of a parallelogram $ABCD$, construct similarly oriented equilateral triangles $ABC_1, BCA_1, CDD_1$, and $DAB_1$. Let $N,P,Q$ and $M$ be the centers of the triangles, respectively. Show that the quadrilateral $MNPQ$ happens to be a parallelogram.

I think this is a really interesting problem. I found it in the geometric transformations unit in my textbook. The given hint is to use the (central) symmetry with center the intersection of the diagonals of the given parallelogram ($ABCD$). I am not sure I see it though. Any help would be appreciated. Thank you in advance!

Answer 1

Let $ R(X)$ be the $180^\circ$ rotation of the point $X$ about the center of parallelogram $ABCD$.

Show that

• $R(A) = C, R(B) = D.$
• $R(C_1) = D_1, R(A_1) = B_1$. -> Note this isn't as symmetric as the rest because of the naming in the question. If we used cyclic notation like $ ABC_1, CDA_1$, then we would have $ R(C_1) = A_1$. In fact, $R(ABC_1) = R(CDA_1)$.
• $R(N) = Q, R(P) = M$.
• $NQ, PM$ bisect each other about the center,
• These points form a parallelogram.

Answer 2

In the above picture, we can use three pairs of congruent triangles to prove the conclusion.

1.For $\triangle MDQ$ $ and $ $\triangle PBN$,

$\alpha=\alpha'$, MD=BP,NB=DQ, $\triangle MDQ$ $\equiv$ $\triangle PBN$ (SAS), $\to$ MQ=NP (1)

2.For $\triangle MAN$ and $\triangle PCQ$,

$\beta=\beta'$,MA=PC,AN=QC,$\triangle MAN$ $\equiv$ $\triangle PCQ$ (SAS), $\to$ MN=PQ (2)

3.For $\triangle MNP$ and $\triangle MPQ$, bb MP=MP ,MQ=NP and MN=PQ from (1)(2), $\triangle MNP$$\equiv$ $\triangle MPQ$ (SSS) $\to$

$\angle NMP=\angle QMP$, NP//MQ, with NP=MQ from (1), MNPQ is a parallelogram.

Answer 3

Here is a proof via the complex geometry. Given the vertexes of the parallelogram at $z_1$, $z_2$, $-z_1$ and $-z_2$, the third vertexes of the equilateral triangles are

$$A = z_1e^{-i\frac\pi3} - z_2e^{i\frac\pi3}, \>\>\> B = z_2e^{-i\frac\pi3} + z_1e^{i\frac\pi3}, \>\>\> C = -z_1e^{-i\frac\pi3} + z_2e^{i\frac\pi3}, \>\>\> D = -z_2e^{-i\frac\pi3} - z_1e^{i\frac\pi3} $$

and their centers $$X = \frac13(A+ z_1 -z_2) = \frac13 z_1(1+e^{-i\frac\pi3}) - \frac13z_2(1+e^{i\frac\pi3})$$ $$Y = \frac13(B+ z_2 +z_1) = \frac13 z_2(1+e^{-i\frac\pi3}) + \frac13z_1(1+e^{i\frac\pi3})$$ $$Z = \frac13(C- z_1 +z_2) = -\frac13 z_1(1+e^{-i\frac\pi3}) + \frac13z_2(1+e^{i\frac\pi3})$$ $$W = \frac13(D- z_2 -z_1) = -\frac13 z_2(1+e^{-i\frac\pi3}) - \frac13z_1(1+e^{i\frac\pi3})$$

Then, it is straightforward to verify

$$Y-X=Z-W,\>\>\>\>\>Z-Y = W-X$$

Thus, $XYZW$ is a parallelogram .

Answer 4

Yet an "other" proof. The following picture...

introduces slightly different notations to have a simpler book keeping (and typing). Here, $O$ is the intersection of diagonals in the given parallelogram $ABCD$, the equilateral triangle constructed on $AB$ has the third vertex $X_{AB}$ and center $O_{AB}$, and similar notations apply to the other ones.

From the congruence (written as equality) $\Delta DAO_{DA}=\Delta BCO_{BC}$ of isosceles triangles (case ASA on bases $AD=BC$) we get $AO_{DA}=CO_{BC}$. Together with $OA=OC$ we obtain $$ \tag{$*$} \Delta OAO_{DA}=\Delta OCO_{BC}\ , $$ since the angle between the pairs of corresponding sides also coincide. Then $(*)$ implies

• $\widehat{ AOO_{DA}}=\widehat{ COO_{BC}}$, and from the colinearity of $A,O,C$ we obtain the colinearity of $O_{DA},O,O_{BC}$,
• $OO_{DA}=OO_{BC}$.

Similar properties can be extrapolated "from the violet pieces to the brown pieces" in the figure.

So $O$ is the mid point of both segments $O_{DA}O_{BC}$ and $O_{AB}O_{CD}$, hence the quadrilateral $\color{blue}{O_{AB}O_{BC}O_{CD}O_{DA}}$ is a parallelogram.

(To obtain $O_{AB}O_{DA}=O_{BC}O_{CD}$ compare the two triangles with these segments as corresponding sides, and with the common vertex in $O$.)

$\square$

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Note: The above is the long story for the proof using the reflection w.r.t. the center $O$ of $ABCD$. This reflection exchanges opposite sides of the parallelogram, thus also opposite constructed equilateral triangles, thus also their centers, and we land immediately at the above displayed property for $\color{blue}{O_{AB}O_{BC}O_{CD}O_{DA}}$, which concludes the proof.