G.105 distances to a circle from an eccentric diameter point — Heath

III.7

If on the diameter of a circle a point be taken which is not the centre of the circle, and from the point straight lines fall upon the circle, that will be greatest on which the centre is, the remainder of the same diameter will be least, and of the rest the nearer to the straight line through the centre is always greater than the more remote, and only two equal straight lines will fall from the point on the circle, one on each side of the least straight line.

Let ( A B C D ) be a circle, and let / A D / be a diameter of it; on / A D / let a point F be taken which is not the centre of the circle, let E be the centre of the circle, and from F let straight lines / F B /, / F C /, / F G / fall upon the circle ( A B C D );

I say that / F A / is greatest, / F D / is least, and of the rest / F B / is greater than / F C /, and / F C / than / F G /.

For let / B E /, / C E /, / G E / be joined.

Then, since in any triangle two sides are greater than the remaining one, [G.45]

/ E B /, / E F / are greater than / B F /.

But / A E / is equal to / B E /;

therefore / A F / is greater than / B F /.

Again, since / B E / is equal to / C E /, and / F E / is common, the two sides / B E /, / E F / are equal to the two sides / C E /, / E F /. But the angle < B E F ) is also greater than the angle < C E F );

[G.49]therefore the base / B F / is greater than the base / C F /.

For the same reason

/ C F / is also greater than / F G /.

Again, since / G F /, / F E / are greater than / E G /, and / E G / is equal to / E D /,

/ G F /, / F E / are greater than / E D /.

Let / E F / be subtracted from each;

therefore the remainder / G F / is greater than the remainder / F D /.

Therefore / F A / is greatest, / F D / is least, and / F B / is greater than / F C /, and / F C / than / F G /.

I say also that from the point F only two equal straight lines will fall on the circle ( A B C D ), one on each side of the least / F D /.

For on the straight line / E F /, and at the point E on it, let the angle < F E H ) be constructed equal to the angle < G E F ) [G.48], and let / F H / be joined.

Then, since / G E / is equal to / E H /, and / E F / is common,

the two sides / G E /, / E F / are equal to the two sides / H E /, / E F /;

and the angle < G E F ) is equal to the angle < H E F );

[G.24]therefore the base / F G / is equal to the base / F H /.

I say again that another straight line equal to / F G / will not fall on the circle from the point F.

For, if possible, let / F K / so fall.

Then, since / F K / is equal to / F G /, and / F H / to / F G /,

/ F K / is also equal to / F H /,

the nearer to the straight line through the centre being thus equal to the more remote: which is impossible.

Therefore another straight line equal to / G F / will not fall from the point F upon the circle;

therefore only one straight line will so fall.

Therefore etc. Q. E. D.

Dependency Graph