Relative Lengths of Lines Outside Circle

Theorem

 * If a point is taken outside a circle and from the point straight lines are drawn through to the circle, one of which is through the center and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the center is greatest, while of the rest the nearer to that through the center is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote; and only two equal straight lines fall on the circle from the point, one on each side of the least.

Proof

 * Euclid-III-8.1.png

Let $ABC$ be a circle and let $D$ be a point outside the circle.

Let $M$ be the center of the circle and let $E$, $F$, and $C$ be points on the concave circumference.

Draw straight lines $DM$, $DC$, $DE$, and $DF$, then extend $DM$ to $A$ on the circle.

Place $H$, $L$, $K$, and $G$ on the intersections of the circle 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 $DE > DF > DC$, and of the straight lines falling on the convex circumference $HLKG$, the line $DG$ is the least, with $DK < DL < DH$.

The proof is as follows:

Join $ME$, $MF$, $MC$, $MH$, $ML$, and $MK$.

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

But we know that $EM + MD > ED$, so it follows that $AD > ED$. Since this is true for any $E$, $AD$ is the greatest line from $D$ to the concave circumference.

Since $EM = FM$, $DM$ is common, and $\angle EMD > \angle FMD$, it follows that $ED > FD$.

Similarly we can show that $FD > CD$. Thus $DA > DE > DF > DC$.

We know that $MK + KD > MD$, and since $MG = MK$, we have $KD > GD$. Since this is true for any $K$, $GD$ is the least line from $D$ to the convex circumference.

Since $K$ is inside $\triangle MLD$, it follows that $ML + LD > MK + KD$. Then since $MK = ML$, we have $DK < DL$.

Similarly we can show that $DL < DH$. Thus $DG < DK < DL < DH$.



Also, from the point $D$ only two equal straight lines fall on the circle $ABC$, one on each side of the line $DA$.

Construct the angle $\angle DMB = \angle KMD$ on the straight line $MD$ at the point $M$. Join $DB$.

Then since $MK = MB$ and $MD$ is common, from Triangle Side-Angle-Side Equality $DK = DB$.

Another straight line equal to $DK$ will not fall on the circle 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 the circle), a contradiction.

Therefore the point $N$ cannot exist as described, and no more than two equal straight lines fall on the circle from $D$, one on each side of $DA$.