Relative Lengths of Lines Inside Circle

Proof

 * Euclid-III-7.png

Let $ABCE$ be a circle, and let $AE$ be a diameter of it.

Let $D$ be the center of the circle, and let $F$ be a point on $AE$ different from $D$.

Let $FB, FC, FG$ be drawn from $F$ to the edge of the circle.

Then $FA$ is the greatest, $FE$ is the least, and (according to the diagram above) of the others, $FB$ is greater than $FC$, while $FC$ is greater than $FG$.

The proof is as follows.

Join $BD, CD, GD$.

From Sum of Two Sides of Triangle Greater than Third Side, we have that $DB + DF > BF$.

But $AD = BD$, so $AF > BF$.

Since $BD = CD$ and $FD$ is common, $BD + DF = CD + CF$.

But $\angle BDF > \angle CDF$, so from the Hinge Theorem $BF > CF$.

For the same reason, $CF > GF$.

Again, since $GF + FD > DG$, and $DG = DE$, $GF + FD > DE$.

Subtract $FD$ from each, and we see that $GF > FE$.

Therefore $FA$ is the greatest, followed by, in order of size, $FB, FC, FG$ and finally $FE$ is the smallest.

Also, from the point $F$ only two equal straight lines fall on the circle $ABCE$, one on either side of the least line $FE$.

Construct the angle $\angle FDH$ equal to $\angle GDF$ on the opposite side of $DF$, and join $FH$.

Since $GD = DH$ and $DF$ is common, and $\angle FDH = \angle GDF$ from Triangle Side-Angle-Side Equality, $\triangle FDH = \triangle GDF$, and so $FG = FH$.

Another straight line equal to $FG$ will not fall on the circle from $F$.

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

Since $FK = FG$, and $FH = FG$, it follows that $FK = FH$.

But this contradicts what we proved earlier.

Therefore only one such straight line can be constructed.