G.143 intersecting chords theorem — Heath

III.35

If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.

For in the circle ( A B C D ) let the two straight lines / A C /, / B D / cut one another at the point E; I say that the rectangle contained by / A E /, / E C / is equal to the rectangle contained by / D E /, / E B /.

If now / A C /, / B D / are through the centre, so that E is the centre of the circle ( A B C D ), it is manifest that, / A E /, / E C /, / D E /, / E B / being equal, the rectangle contained by / A E /, / E C / is also equal to the rectangle contained by / D E /, / E B /.

Next let / A C /, / D B / not be through the centre; let the centre of ( A B C D ) be taken, and let it be F; from F let / F G /, / F H / be drawn perpendicular to the straight lines / A C /, / D B /, and let / F B /, / F C /, / F E / be joined.

Then, since a straight line / G F / through the centre cuts a straight line / A C / not through the centre at right angles,

[G.100]it also bisects it;

therefore / A G / is equal to / G C /.

Since, then, the straight line / A C / has been cut into equal parts at G and into unequal parts at E, the rectangle contained by / A E /, / E C / together with the square on / E G / is equal to the square on / G C /; [G.87]

Let the square on / G F / be added; therefore the rectangle / A E /, / E C / together with the squares on / G E /, / G F / is equal to the squares on / C G /, / G F /.

But the square on / F E / is equal to the squares on / E G /, / G F /, and the square on / F C / is equal to the squares on / C G /, / G F /; [G.79] therefore the rectangle / A E /, / E C / together with the square on / F E / is equal to the square on / F C /.

And / F C / is equal to / F B /; therefore the rectangle / A E /, / E C / together with the square on / E F / is equal to the square on / F B /.

For the same reason, also, the rectangle / D E /, / E B / together with the square on / F E / is equal to the square on / F B /.

But the rectangle / A E /, / E C / together with the square on / F E / was also proved equal to the square on / F B /; therefore the rectangle / A E /, / E C / together with the square on / F E / is equal to the rectangle / D E /, / E B / together with the square on / F E /.

Let the square on / F E / be subtracted from each; therefore the rectangle contained by / A E /, / E C / which remains is equal to the rectangle contained by / D E /, / E B /.

Therefore etc.

Dependency Graph