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Showing that equilateral pentagon with three adjacent equal angles is regular using only euclidean geometry.

Hypothesis: sides $AB=BC=CD=DE=EF$ and angles $\angle EAB=\angle ABC=\angle BCD.$
Thesis: angles $\angle ABC=\angle BCD=\angle CDE=\angle DEA=\angle EAB.$

For simplicity we call $EAB=ABC=BCD:=\gamma$. Using congruence criteria of triangles, we have
$BEA=CAB=ABE=DBC=BCA=CDB:=\alpha$,
$BMA=CNB=\gamma$,
$\gamma+2\alpha=\pi$,\ $AME=CMB=BNA=DNC=2\alpha$,
$BED=EDB=2\alpha$,
$DEA=CDE=3\alpha$.\

Now, the goal is to show that $3\alpha=\gamma$. I don't know how to continue from here or other ways to solve this exercise using euclidean geometry. Thanks, any help is appreciated.

enter image description here

Piita
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Jess
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  • Have you drawn a diagram? – Mark Bennet Jun 08 '23 at 21:15
  • One possible approach is to start out with $AB=BC=CD=DE = x$ and $\angle ABC = \angle BCD = \angle CDE = \alpha$, and calculate $EA$ in terms of $x$ and $\alpha$ using trigo. Then, show that $ EA = x$ iff $ \alpha = 108^\circ$ – Calvin Lin Jun 08 '23 at 21:25

2 Answers2

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Here is another approach:

enter image description here

In the figure, draw the angle bisectors of $\angle B$ and $\angle C$. Let them meet at $O$. Join the other lines as shown.

Let $\angle EAB =\angle ABC = \angle BCD = x.$

Note that

$(1)$ $OC=OB, \angle OCB=\angle OBA$ and $CB=BA$

$(2)$ $\therefore \Delta OBA \cong \Delta OCB$ (SAS)

$(3)$ Similarly we can prove that all five triangles in the figure are congruent.

$(4)$ $\therefore \angle A=\angle B=\angle C=\angle D=\angle E=x.$

$(5)$ Since $AB=BC=CD=DE=EA$ is given.

$(6)$ $ABCDE$ is a regular pentagon.

Li Kwok Keung
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I apologize in advance if I made any mistakes, but I think I found a proof.

My diagram

$\triangle EAB \cong \triangle DCB$ (SAS congruence), so $\overline{EB} \cong \overline{DB}$. $\triangle BED$ is isosceles.

$\therefore$ $\angle BED \cong \angle BDE$. $\angle BEA \cong \angle BDC$.

$\implies$ $\angle AED\ \cong \angle CDE$.

Switching focus now:

$\triangle DCB \cong \triangle CBA$ by SAS congruence. $\overline{DB} \cong \overline{CA}$ because CPCTC. $\overline{AD} \parallel \overline{BC}$ because ABCD is a square, rectangle, or isosceles trapezoid, all of which have parallel opposite sides.

In a similar manner: $\triangle CBA \cong \triangle BAE$ by SAS. $\overline{CA} \cong \overline{BE}$ by CPCTC. $\overline{CE} \parallel \overline{BA}$ for the same reason as before.

By carrying out the same process, we can prove that all of the diagonals are parallel to their corresponding side.

Angle chasing leads to the fact that the diagonals of the large pentagon trisect the larger angle, and each of the trisected angles are congruent. Therefore, the pentagon is equiangular and equilateral, which makes it regular.

Capt
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