280 Strong’s Problems. 
Through D draw DE perpendicular to AB, and extend if 
til EG=DE. Through D, G describe a circle IDG 
touching LIO in I. The centre of this circle is in the line 
Demonstration. For join FD, FC. Now the line FC 
will pass through I the point of contact of the circles LIO, 
IGD. Th 
Cl=z and FI=FD; therefore FC—FD=a, as was re- 
quired. 
Prosiem XIV. 
There are two points and a straight line given im position, 
it is required to find a point in the straight line, such that 
the sums of the lines drawn from given points to this point 
shall be equal to a given line, this line never being less thar 
the line joining the two points. : 
Const. Let (Fig. 15. and 16. pl.3.) AB be the given line 
and C, D the given points it is required to find a point in the 
given line such that the lines drawn from the given pots 
to that point shall together be equal to a given line. Draw 
DH at right angles to AB and extend it, till HF =DH. 
From C with radius=the given sum describe the circle EB 
1. Through (P. IlI,) D, E describe the circle DEF touch- 
ing EBI in E. Now because DF is bisected at right an- 
gles by AB, the centre of DEF falls in BA. Let G be the 
centre then G is the point required. 
Demonstration. Yor join CG which extended will pass 
through the point of contact of the circles DEF, BEI. 
Join also GD. Now CE=CE+GE=CG+GD. But 
CE=the given line. Therefore CG-+GD=the given line 
as was required. 
