Strong’s Problems. : 
Demonstration. Because the angle FCD = angle GCD 
and the angle DF =angle DGC, and DC is common to 
both the triangles DFC, DGC, the straight line DF — 
straight line DG. In like manner it may be shewn that 
DG=DE. Therefore a circle described from D as cen- 
tre with DF as radius will pass through the three points 1, 
F, G. And it is manifest also that it touches the lines BH, 
AC, CB, in those points, since the radii DE, DG, DF are 
severally perpendicular to the lines BH, BC, CA.-Q. ELL. 
Prosuem IIf. 
Given two points and a straight line in position, the 
points not being on opposite sides of the line ; it is required 
to describe a circle the circumference of which shall pass 
through the two given points, and touch the given line. — 
Case I. When one of the given points is in the given 
straight line. / 
Const. Let AB (Fig. 3.) be the given straight line, C 
the given point in AB, and D the other given point.—Join 
DC, and through C draw CE at right angles to AB. At 
the point D in the line DC, make the angle CDE = the an- 
gle DCE. Then the side DE = side CE. Therefore a cir- 
cle described from E. as a centre with radius DE, will pass 
through C, and D. And it will likewise ‘touch the line 
AB, this line being perpendicular to the radius CE. 
Case II. When the straight line joining the two given 
points is parallel to the given straight line. 
Const. Let AB, (Fig. 4.) be the given straight line, and 
C, D, the two given points. Join CD; and bisect CD in 
F. From F draw FE, at right anglestoCD. Let FE ex- 
tended cut ABin E. Through C, D, E describe a circle, 
which shall be the circle required. 
Demonst. For Jom ED, and EC. Because the angles 
EFC, EFD are equal, and CF = FD and FE is common, 
the angle FCE — angle FDE. But the angle FCE — 
alternate angle CEA. Therefore CEA = CDE. Therefore 
AB touches the circle CDE in the point E. (Eucl. III. 22.) 
Case HI. When the straight line joining the two given 
points is oblique io the given line. 
Const. Jom CD (Mig. 5.) and let CD produced meet AB in 
B. Take BE = a mean proportional between BD and BC. 
