Pentagon $ABCDE$ is given inside a circle radius $r$. If $AB=BC=DE=r$, prove that $BGF$ is an equilateral triangle where $G$ and $F$ are...

Question

Pentagon $ABCDE$ is given inside a circle radius $r$. If $AB=BC=DE=r$, prove that $BGF$ is an equilateral triangle where $G$ and $F$ are midpoints of sides $CD$ and $EA$ of the pentagon $ABCDE$.

I've went ahead and drawn up a sketch on GeoGebra, I apologize for its' rough edges as it's my first time using the program. It's easy to calculate all of the angles around the center $(AIE,EID,DIC,CIB,BIA),$ but I haven't gotten anywhere doing this. I've tried adding new points to find some cyclic quadrilaterals but nothing's worked as the triangle's in a really awkward position. If anyone has an idea, please feel free to share it as I've been staring at a sketch of this problem for over an hour. Thanks in advance.

Answer 1

Here is a proof using complex numbers.

Without loss of generality let the radius of the circle be $1$. Let the origin $O$ be the centre of the circle and let $\overrightarrow{OC}=1$.

Let $\angle AOE=\theta$.

Let $\omega$ be a cube root of unity, so that $\omega^3=1$ and $1+\omega+\omega^2=0$

Since multiplying a vector by $\omega$ corresponds to a rotation of that vector by $120^o$ anticlockwise, it suffices to show that $$\omega\overrightarrow{GB}=\overrightarrow{BF}$$

We have:$$\overrightarrow{OA}=\omega$$ $$\overrightarrow{OB}=1+\omega$$ $$\overrightarrow{OE}=\omega e^{i\theta}$$ $$\overrightarrow{OD}=(1+\omega)\omega e^{i\theta}$$ $$\overrightarrow{OF}=\frac12(\omega+\omega e^{i\theta})$$ $$\overrightarrow{OG}=\frac12(1+(1+\omega)\omega e^{i\theta})$$

Then, $$\overrightarrow{GB}=(1+\omega)-\frac12(1+(1+\omega)\omega e^{i\theta})$$ $$\implies \omega\overrightarrow{GB}=\omega+\omega^2-\frac12(\omega+(\omega^2+\omega^3)e^{i\theta}$$ $$=\frac12(\omega+2\omega^2-\omega^2e^{i\theta}-e^{i\theta})$$

Meanwhile, $$\overrightarrow{BF}=\frac12(\omega+\omega e^{i\theta})-(1+\omega)$$

Therefore, $$\omega\overrightarrow{GB}-\overrightarrow{BF}=\frac12(\omega+2\omega^2-\omega^2 e^{i\theta}-e^{i\theta}-\omega-\omega e^{i\theta}+2+2\omega)$$

$$=\frac12(2(1+\omega+\omega^2)-e^{i\theta}(1+\omega+\omega^2))=0$$

Hence proved.

Answer 2

Here is a path

1. Let $O$ be the circle center. Show that $\angle AOE$ and $\angle COD$ are supplementary angles. Then $\angle AOF$ and $\angle COG$ are complementary angles, and $$\triangle AOF \cong \triangle COG$$by ASA Congruence Criterion.
2. Using 1. and SAS Congruence Criterion show that $\triangle ABF \cong \triangle OBG$.
3. Conclude that $\angle ABF \cong \angle OBG$ and therefore $\measuredangle FBG = 60^\circ$. Also $BF \cong BG$, hence $\triangle BFG$ is equilateral. $\blacksquare$

Answer 3

First solution: We are using complex numbers. Let $z=\zeta_6=\exp\frac{2\pi i}6=\frac 12(1+i\sqrt 3)$ be this chosen primitive root of order six. So $z^2-z+1=0$, and $z^3=-1$.

Translate, rescale and rotate the given configuration of points, so that the circle $(ABCDE)$ becomes the unit circle, and $C$ gets the affix $c=1\in\Bbb C$. In general, for a point variable we denote by the lower case variable its corresponding affix in $\Bbb C$. Then for some $u$ with modulus one the points $C,B,A,E,D$ are $c=1$, $b=z$, $a=z^2$, $e=z^2u$, $d=z^3u$.

Let us check that a $60^\circ$-rotation around $B$ brings $F$ into $G$, i.e. that $z(f-b)$ and $(g-b)$ are equal. $$ \begin{aligned} 2z(f-b) &= 2z\left(\frac 12(a+e)-b\right)=(z^3 + z^3u)-2z^2 =-1-u-2(z-1)=1-u-2z\ ,\\ 2(g-b) &= 2\left(\frac 12(c+d)-b\right)=(1+z^3u)-2z=1-u-2z\ . \end{aligned} $$ $\square$

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Second and same solution. We are doing exactly the same as above, translating algebra into geometry. Let $H$ be the reflection in $G$ of $O$, the circumcenter of $ABCDE$.

Then the isosceles triangles $\Delta OAE$ and $\Delta COH$ have the same sides $OA=OE=CO=CH$, and their angles in $O$, respectively $C$ are equal. (Since adding the same angle $\sphericalangle COD$ to each gives $180^\circ$, the sums being computed in the rhombus $CODH$, respectively around the center $O$ of the circle $(ABCDE)$.)

Consider now a $60^\circ$-rotation around $B$. The equilateral triangle $\Delta BAO$ is rotated into $\Delta BOC$. The "assembled" puzzle built out of the two pieces $(\Delta BAO,\Delta OAE)$ goes to $(\Delta BOC, \Delta COH)$. So the mid point $F$ of $AE$ goes to the mid point $G$ of $CD$.

So $\Delta BFG$ has $BF=BG$ and the angle in $B$ is $60^\circ$, i.e. it is equilateral.

$\square$