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

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 :
 * $AC = BE$

and:
 * $\triangle ABC = \triangle ABE$

Therefore:
 * $\triangle BAC = \triangle ABE$

From :
 * $\angle AHE = 2 \cdot \angle BAH$

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

Therefore from:

and:

it follows that:
 * $\angle EAC = 2 \cdot \angle BAC$

Therefore:
 * $\angle HAE = \angle AHE$

Hence from :
 * $HE = EA$

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 :
 * $\angle BAE = \angle AHB$

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

Therefore from :
 * $EB : BA = AB : BH$

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 $ABCDE$.