If a straight line be cut in extreme and mean ratio, the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment.
For let any straight line / A B / be cut in extreme and mean ratio at the pointC, let / A C / be the greater segment, and let / A C / be bisected at D ; I say that the square on / B D / is five times the square on / D C /.
For let the square < A E > be described on / A B /, and let the figure be drawn double.
Since / A C / is double of / D C /,
therefore the square on / A C / is quadruple of the square on / D C /, that is, < R S > is quadruple of < F G >.
And, since the rectangle/ A B /, / B C / is equal to the square on / A C /, and < C E > is the rectangle / A B /, / B C /,
therefore < C E > is equal to < R S >.
But < R S > is quadruple of < F G > ;
therefore < C E > is also quadruple of < F G >.
Again, since / A D / is equal to / D C /,
/ H K / is also equal to / K F /.
Hence the square < G F > is also equal to the square < H L >.
Therefore / G K / is equal to / K L /, that is, / M N / to / N E / ; hence < M F > is also equal to < F E >.
But < M F > is equal to < C G > ;
therefore < C G > is also equal to < F E >.
Let < C N > be added to each ;
therefore the gnomon< O P Q > is equal to < C E >.
But < C E > was proved quadruple of < G F > ;
therefore the gnomon < O P Q > is also quadruple of the square < F G >.
Therefore the gnomon < O P Q > and the square < F G > are five times < F G >.
But the gnomon < O P Q > and the square < F G > are the square < D N >.
And < D N > is the square on / D B /, and < G F > the square on / D C /.
Therefore the square on / D B / is five times the square on / D C /.
