424 A Demonstration of 
angle hMg = the supplement of hGg, and 
consequently of hE’E ; therefore hLE = 
hMeg, Mg parallel to LE, andhM: ML: : 
he :gE::Gh’:EG:: Fh: FE. Which 
is Proposition tenth, part 2d. ; 
At F erect a perpendicular to FH, 
produce Hh till it meets it in 3, and join 
Dh; then the z DhH being right, F, 3, 
h, D are in a circle; therefore Hh . H3 = 
FH. DH—FH’?—FD.FH —H3*—H. 
th; .*. Hy. 9h — He’— FH’ + FD. FH 
1 =Fy 4+ FD .FH. Which is the eleventh 
Proposition. 
If Fd’ = Fa = Fq = FT, then 32= Fq 
+ F3, ad’ = Fq — Fa, 3a* + ad — 2Fq* 
equal to the rectangle EGF; then I say the square of CD 
will be equal to the rectangle ACB. 
Prop. XIV. Things remaining as inthe last proposition, 
if the perpendiculars Eg an@ FH be.demitted ; then I say 
that the rectangle gCH is equal to the square of CD. 
Prop. XV. If from C any point in the diameter of a 
circle AB produced a tangent be drawn, and from the 
point of contact Da perpendicular to the diameter DE be 
demitted; then I say that AC :CB:: AE: EB. 
Or conversely thus : 
If in AB the diameter of a circle be taken two points C 
and E such that AC: CB:: AE: EB, and from Ea per-— 
pendicular ED raised, and CD ate then I say CD 
touches the circle in D. 
Or thus: 
