Straight Lines Subtending Two Consecutive Angles in Regular Pentagon cut in Extreme and Mean Ratio

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Theorem

In the words of Euclid:

If in an equilateral and equiangular pentagon straight lines subtend two internal angles taken in order, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon.

(The Elements: Book $\text{XIII}$: Proposition $8$)


Proof

Euclid-XIII-8.png

Let $ABCDE$ be a regular pentagon.

Let the straight lines $AC$ and $BE$ subtend two vertices $A$ and $B$ taken in order.

Let $AC$ and $BE$ cut each other at $H$.

It is to be demonstrated that $AC$ and $BE$ are cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$.


Let the circle $ABCDE$ circumscribe the pentagon $ABCDE$.

We have that:

$EA = AB$

and:

$AB = BC$

and they contain equal angles.

Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$AC = BE$

and:

$\triangle ABC = \triangle ABE$

Therefore:

$\triangle BAC = \triangle ABE$

From Proposition $31$ of Book $\text{I} $: Sum of Angles of Triangle equals Two Right Angles:

$\angle AHE = 2 \cdot \angle BAH$

We have that the arc $EDC$ is twice the arc $DC$.

Therefore from:

Proposition $28$ of Book $\text{III} $: Straight Lines Cut Off Equal Arcs in Equal Circles

and:

Proposition $33$ of Book $\text{VI} $: Angles in Circles have Same Ratio as Arcs

it follows that:

$\angle EAC = 2 \cdot \angle BAC$

Therefore:

$\angle HAE = \angle AHE$

Hence from Proposition $6$ of Book $\text{I} $: Triangle with Two Equal Angles is Isosceles:

$HE = EA$

That is:

$HE = AB$

We have that:

$BA = AE$

and:

$\angle ABE = \angle AEB$

But it has been proved that:

$\angle ABE = \angle BAH$

Therefore:

$\angle BEA = \angle BAH$

We also have that $\angle ABE$ is common to $\triangle ABE$ and $\triangle ABH$.

Therefore from Proposition $32$ of Book $\text{I} $: Triangle with Two Equal Angles is Isosceles:

$\angle BAE = \angle AHB$

Therefore $\triangle ABE$ has the same corresponding angles with $\triangle ABH$.

Therefore from Proposition $4$ of Book $\text{VI} $: Equiangular Triangles are Similar:

$EB : BA = AB : BH$

But:

$BA = EH$

Therefore:

$BE : EH = EH : HB$

and:

$BE > EH$

Therefore:

$EH > HB$

Therefore $BE$ is cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$.

Similarly it is shown that $AC$ is also cut in extreme and mean ratio at the point $H$, and that the greater segment equals the side of $ABCDE$.

$\blacksquare$


Historical Note

This proof is Proposition $8$ of Book $\text{XIII}$ of Euclid's The Elements.


Sources

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