G.93 cut a segment in extreme and mean ratio — Heath

II.11

To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

Let / A B / be the given straight line;

thus it is required to cut / A B / so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

For let the square < A B D C > be described on / A B /; [G.75]

let / A C / be bisected at the point E, and let / B E / be joined; let / C A / be drawn through to F, and let / E F / be made equal to / B E /; let the square < F H > be described on / A F /, and let / G H / be drawn through to K.

I say that / A B / has been cut at H so as to make the rectangle contained by / A B /, / B H / equal to the square on / A H /.

For, since the straight line / A C / has been bisected at E, and / F A / is added to it,

[G.88]the rectangle contained by / C F /, / F A / together with the square on / A E / is equal to the square on / E F /.

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

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

But the squares on / B A /, / A E / are equal to the square on / E B /, for the angle at < B A E ) is right: [G.79]

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

Let the square on / A E / be subtracted from each;

therefore the rectangle / C F /, / F A / which remains is equal to the square on / A B /.

Now the rectangle / C F /, / F A / is < C F G K >, for / A F / is equal to / F G /; and the square on / A B / is < A B D C >;

therefore < C F G K > is equal to < A B D C >.

Let < A H K C > be subtracted from each;

therefore < A F G H > which remains is equal to < H B D K >.

And < H B D K > is the rectangle / A B /, / B H /, for / A B / is equal to / B D /; and < A F G H > is the square on / A H /;

therefore the rectangle contained by / A B /, / B H / is equal to the square on / H A /.

therefore the given straight line / A B / has been cut at H so as to make the rectangle contained by / A B /, / B H / equal to the square on / H A /.

Dependency Graph

Required for

G.162, G.163, G.164