PROPOSITION 11. PROBLEM.
To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.
Let AB be the given straight line: it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.
On AB describe the square
ABDC; [I. 46.
bisect AC at E; [I. 10.
join BE; produce CA to F, and
make EF equal to EB; [I. 3.
and on AF describe the square
AFGH. [1. 46.
AB shall be divided at H so
that the rectangle AB, BH is
equal to the square on AH.
Produce GH to K.
Then, because the straight line
AC is bisected at E, and pro-
duced to F, the rectangle CF, FA, together with the
square on AE, is equal to the square on EF. [II. 6.
But EF is equal to EB. [Construction.
Therefore the rectangle CF, FA, together with the square
on AE, is equal to the square on EB.
But the square on EB is equal to the squares on AE,AB,
because the angle EAB is a right angle. [I. 47.
Therefore the rectangle CF, FA, together with the square
on AE, is equal to the squares on AE, AB.
Take away the square on AE, which is common to both;
therefore the remainder, the rectangle CF, FA, is equal to
the square on AB. [Axiom 3.
But the figure FK is the rectangle contained by CF, FA,
for FG is equal to FA;
and AD is the square on AB;
therefore FK is equal to AD.
Take away the common part AK, and the remainder FH
is equal to the remainder HD. [Axiom 3.
