unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Let the straight line AB be divided into two equal parts at the point C, and into two unequal parts at the point D : the rectangle AD,DB, together with the square on CD, shall be equal to the square on CB.
On CB describe the
square CEFB ; [l- 46.
join BE; through D draw
DHG parallel to CE or BF;
through H draw KLM paral-
lel to CB or EF ; and through
A draw AK parallel to CL
or BM. [I. 31.
Then the complement CH is equal to the complement HF, [I. 43.
to each of these add DM therefore the whole CM is equal
to the whole DF. [Axiom 2.
But CM is equal to AL, [I. 36.
because AC is equal to CB. [Hypothesis.
Therefore also AL is equal to DF, [Axiom 1.
To each of these add CH; therefore the whole AH is equal
to DF and CH. [Axiom 2.
But AH is the rectangle contained by AD, DB, for DH is equal to DB ; [II. 4, Corollary.
and DF together with CH is the gnomon CMG ;
therefore the gnomon CMG is equal to the rectangle AD,DB
To each of these add LG, which is equal to the square on
CD. [II. 4, Corollary, and I. 34.
Therefore the gnomon CMG, together with LG, is equal to
the rectangle AD,DB, together with the square on CD. [Ax.2.
But the gnomon CMG and LG make up the whole figure
CEFB, which is the square on CB.
Therefore the rectangle AD,DB, together with the square on CD, is equal to the square on CB.
Wherefore, if a straight line &c. q.e.d.
Prom this proposition it is manifest that the difference of the squares on two unequal straight lines AC, CD, is equal to the rectangle contained by their sum and difference.
