250 Strong’s Problems. 
Through D draw DE perpendicular to AB, and extend it 
iil EG=DE. Through D, G describe a circle IDG 
touching LIO in I. The centre of this circle is in the line 
AB, for the chord DG is bisected at right angles by AB. 
Let F be the centre. Then F is the point to be found. 
Demonstration. For join FD, FC. Now the line FC 
will pass through I the point of contact of the circles LIO, 
(GD. Then, FC=FI+IC, and FC—ED=CI. But 
Cl=zr and FI=FD; therefore FC—FD=a, as was re- 
quired. 
Prosiem XIV. 
There are two points and a straight line given in 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 lme never being less than 
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 points 
to that point shall together be equal to a given line. Draw 
DH at right angles to AB and extend it, till HE =DH. 
From C with radius=the given sum describe the circle EB 
{. Through (P. III) 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. $e, 
Demonstration. For join CG which extended will pass 
through the point of contact of the circles DEF, BEL. 
Join also GD. Now CE=CE+GE=CG+GD. But 
CE=the given line. Therefore CG4+ GD==the given line 
as was required. 
