some Properties of Tangents to Circles ; &c, 3^3 
into two parts whose rectangle shall be equal to the square of 
GH. Make GV equal to the less part, and GD to the greater; 
taking V and D in EG produced beyond G. Through D, 
perpendicular to GD, draw LR the base required : to which V 
is the common vertex. 
Dem. First, A is greater than 2GH. For sq. ES — sq. GH 
is greater than sq. EH — sq. GH, which is * sq. EG + 2EGH; 
whence A is greater than _ eg, which is 2GH. 
Therefore A can be divided as required. 
Through V draw VQ parallel to LR; cutting HFK in F, 
and SOT in Q. Draw GF, FD, EQ, OD. Then, because the 
rectangle DGV is equal to the square of GH or GF, the tri- 
angles DGF and FGV are similar, the right angle GVF is 
equal to GFD, DF is a tangent, and V the vertex -f in HFK 
to the base LR. 
Further, To A (DG + GV) add EG. Then 
is equal to DE -j- GV; whence sq. ES = DEG + EGV + 
sq. GH (DGV), = DGE % + sq. EG + EGV + DGV, = 
DG -J- EG X EG -E GV, = DEV. Since, then, the square 
of ES or EO is equal to DEV, DO (as DF before) is a tan- 
gent, and V the vertex in SOT to LR. Q. E. D, 
EV and ED might have been found, instead of GV and 
GD ; mutatis mutandis : by making equal to 
+ EG, and the rectangle under its parts equal to the square 
of ES. 
* II Eucl. prop. 4. 
f Cor. 1 to prop. I. 
t II Eucl, prop. 3. 
