Let AB in the following diagram be the given straight line. With the
centres A and B and radius AB describe the two circles. Mark off DE and
EF equal to AD. With the centres A and F and radius DF describe arcs
intersecting at G. With the centres A and B and distance BG describe
arcs GHK and N. Make HK equal to AB and HL equal to HB. Then with
centres K and L and radius AB describe arcs intersecting at I. Make BM
equal to BI. Finally, with the centre M and radius MB cut the line in C,
and the point C is the required middle of the line AB. For greater
exactitude you can mark off R from A (as you did M from B), and from R
describe another arc at C. This also solves the problem, to find a point
midway between two given points without the straight line.
[Illustration]
I will put the young geometer in the way of a rigid proof. First prove
that twice the square of the line AB equals the square of the distance
BG, from which it follows that HABN are the four corners of a square. To
prove that I is the centre of this square, draw a line from H to P
through QIB and continue the arc HK to P. Then, conceiving the necessary
lines to be drawn, the angle HKP, being in a semicircle, is a right
angle. Let fall the perpendicular KQ, and by similar triangles, and from
the fact that HKI is an isosceles triangle by the construction, it can
be proved that HI is half of HB. We can similarly prove that C is the
centre of the square of which AIB are three corners.