To place at a given point (as an extremity) a straight line equal to a given straight line.
To place at a given point (as an extremity) a
straight line equal to a given straight line..
Thus it is required to place at the point A
(as an extremity) a straight line equal to the
given straight line BC
From the point A to the point B let the straight
line AB be joined,
and on it let the equilateral triangle DAB be constructed.
Let the straight lines AE, BF be produced in a
straight line with DA, DB;
with centre B and distance BC let the circle
CGH be described
and again, with centre D and distance DG let the
circle GKL be described.
Then, since the point B is the centre of the circle CGH,
(a) BC is equal to BG
Again, since the point D is the centre of the circle GKL,
DL is equal to DG.
And in these DA is equal to DB;
therefore the remainider AL is equal to the remainder BG.
But BC was also proved equal to BG;
therefore each of the straight lines
AL, BC is equal to BG.
And things which are equal to the same thing are also
equal to one another;
Therefore at the given point A the straight line AL
is placed equal to the given straight line BC.
(Being) what it was required
to do.
Q.E.F