Commentary on Proposition XIII.3

Back to Proposition XIII.3.

In this case we have

(sq. on BD) = (sq. FG) + (rect. CG) + (rect. CN) = (sq. FG) + (rect. FE) + (rect. CN) = (sq. FG) + (rect. CE) = (sq. FG) + (rect. AB, BC) = (sq. FG) + (sq. on AC), by hypothesis, = 5 (sq. on DC).

The theorem is still more obvious if the figure of II. 11 be used. Let CF be divided in extreme and mean ratio at E, by the method of II. 11.

B C E F D A G

Then, since

(rect. AB, BC) + (sq. on CD) = sq. on BD = sqs. on CD, CF, (rect. AB, BC) = (sq. on CF) = (sq. on CA),

and AB is divided at C in extreme and mean ratio.

And

(sq. on BD) = (sq. on DF) = 5 (sq. on CD).