372 Dr. R. Hey’s Propositions containing 
Then FX and AY are chords through V in two circles having 
V their common vertex to the base LR; whence YX pro- 
duced will * meet FA in L. And FY and AX cut NO, there- 
fore are not parallel to LR, therefore 'f* not to each other, but 
meet in a point of LR. This they cannot do, unless X be 
at B and Y at C : for, since X and Y are such that YX is 
directed to L, FY and AX will, if not passing through C and 
B, either meet before they cut LR, or beyond it. Therefore 
no two circles, passing through F and A respectively, except 
BFT and ACK, have V their common vertex to LR. Therefore 
those two have. 
Let now AC be conceived § to coincide with NO. Then a 
circle, having its centre any point in DV indefinite, might pass 
through A and C. If a centre be found, in DV, by drawing 
DA or DC, and, perpendicular to it, a line from A or C cut- 
ting DV, then the circle will have V its vertex to LR; because 
DA and DC will be tangents. 
Therefore, whether AC cut NO or not, LR is the base to 
which V is vertex in the circles BFT, ACK. Q. E. D. 
Cor. 1. If FA and CB were parallel; RL parallel to them 
would be the base to which V would be vertex, in two such 
circles. For, let VD be a perpendicular on LR ; and E, G, 
X, Y, as before. Then, if YX could meet FA, it would be (f 
in RL. But RL does not meet FA. Therefore YX does not; 
but is parallel to FA and CB ; whence AX and FY, which 
must ^ meet in LR, must pass through B and C, and there- 
fore meet in R ; whence R is a point of the base to which V 
* Prop. Xf. f Ib; latter part. 
§ This may be conceived without an additional figure. 
II Prop. XI. ^ Ib. 
J Prop. XI. 
