|If a straight line is cut at random, then the square on the
whole equals the sum of the squares on the segments plus twice
the rectangle contained by the segments.| |Let the straight line
AB be cut at random at C. I say that the square on AB equals the
sum of the squares on AC and CB plus twice the rectangle AC by
CB. |Describe the square ADEB on AB. Join BD. Draw CF through C
parallel to either AD or EB, and draw HK through G parallel to
either AB or DE.||I.46 |Then, since CF is parallel to AD, and BD
falls on them, the exterior angle CGB equals the interior and
opposite angle ADB.||I.29| |But the angle ADB equals the angle
ABD, since the side BA also equals AD. Therefore the angle CGB
also equals the angle GBC, so that the side BC also equals the
side CG.||I.5 |But CB equals GK, and CG to KB. Therefore GK also
equals KB. Therefore CGKB is equilateral.||I.34| |I say next
that it is also right-angled.| |Since CG is parallel to BK, the
sum of the angles KBC and GCB equals two right angles.||I.29|
|But the ang