Strong’s Problems. 269 
Az=AC.AD. ThenAD: Az:: Ar: Ac. But Ar > AC 
therefore AD~Az, therefore the problem is possible, for 
the point z always falls between C and D. Having deter- 
mined the points 7, z, use them in the same manner as the 
ints C, D were used, and the solution is the same as be- 
ore. 
Cor. 6. In Case III. when the point is between B and 
D as r, then making the rectangle Ar. Az=AD AC, the 
point z will fall beyond C3; and therefore the problem is 
possible. Using then the points r, z for C, D, the solution 
remains the same. On the contrary when the given point 
is beyond C, r willbe between D, and B. Then proceed 
as before. 
Note.—When the circles do not cut, and one does not 
fall entirely within the other, the point cannot be given 
within one of the circles, but must be without the circle, or 
in the circumference of one of them; and then the solu- 
- tion will fall under one of the above cases ; when the circles 
eut each other, the point may be given any where, except 
at the points of intersection of the circles ; when one of the 
circles falls wholly within the other, the point must be given 
between the circumferences of the two circles; in all which 
cases, the construction may be referred to one of the above 
methods. 
Prospiem X. 
the touching circle comprehends them all, or none of them. 
onst. L 
Demonstration. For, join OAD. Now OAD=OA+ 
Ox. EL... 
