Relative Lengths of Lines Outside Circle

Construction

 * Euclid-III-8-1.png

Let $\bigcirc ABC$ be a circle.

Let $D$ be a point outside $\bigcirc ABC$.

Using, let $M$ be the center of $\bigcirc ABC$.

Let $E$, $F$, and $C$ be points on the concave circumference of $\bigcirc ABC$

Using, draw straight lines $DM$, $DC$, $DE$, and $DF$.

Using, extend $DM$ to $A$ on $\bigcirc ABC$.

Place $H$, $L$, $K$, and $G$ on the intersections of $\bigcirc ABC$ and $DC$, $DF$, $DE$, and $DA$ respectively.

Then:


 * of the straight lines falling on the concave circumference $AEFC$, the line $DA$ through the center is the greatest, with $DE > DF > DC$
 * of the straight lines falling on the convex circumference $HLKG$, the line $DG$ is the least, with $DK < DL < DH$.

Proof
Using, join $ME$, $MF$, $MC$, $MH$, $ML$, and $MK$.

Since $AM = EM$, add $DM$ to each, so $AD = EM + MD$.

But from, we know that $EM + MD > ED$.

Hence it follows that:
 * $AD > ED$

Since this is true for any $E$, $AD$ is the greatest line from $D$ to the concave circumference.

We have that:
 * $EM = FM$
 * $DM$ is common
 * $\angle EMD > \angle FMD$.

Hence, from the, it follows that:
 * $ED > FD$

Similarly we can show that:
 * $FD > CD$

Thus:
 * $DA > DE > DF > DC$

From, we know that:
 * $MK + KD > MD$

Because $MG = MK$, we have:
 * $KD > GD$.

Because $K$ is arbitrary, it follows that $GD$ is the least line from $D$ to the convex circumference.

We have that $K$ is inside $\triangle MLD$.

Hence from :
 * $ML + LD > MK + KD$.

Then because $MK = ML$:
 * $DK < DL$

Similarly we can show that $DL < DH$.

Thus:
 * $DG < DK < DL < DH$



From the point $D$ only two equal straight lines fall on $\bigcirc ABC$: one on each side of the line $DA$.

From, construct the angle $\angle DMB = \angle KMD$ on the straight line $MD$ at the point $M$.

Using, join $DB$.

We have that:
 * $MK = MB$
 * $MD$ is common.

Hence from :
 * $DK = DB$

Another straight line equal to $DK$ will not fall on $\bigcirc ABC$ from $D$.

For if this is possible, let $DN$ be this straight line.

Then $DN = DB$.

But from what we proved above, either:
 * $DN > DB$

or:
 * $DN < DB$

depending on where $N$ falls on $\bigcirc ABC$.

This is a contradiction.

Therefore the point $N$ cannot exist as described.

It follows that no more than two equal straight lines fall on $\bigcirc ABC$ from $D$: one on each side of $DA$.