Lawson's Geometrical Theorems. 433 
= FD.FH. Which is the twenty-sixth 26 
Proposition. 
The 28th is the converse of this, viz. if 
FY .Fa = FD. FH then DG: GH:: 27 
FD: FH. And the 27th of the 24th. 28 
If Fa — FT and aH, ah be drawn cut- 
ting the circle in p and o, then since Fh : 
Fa: : Fa: FE, the triangles Fha, FaH 
are equiangular, therefore the zFha = 
Fak, also ao: ap :: aE : ah; therefore 
the ‘triangles aop, ahE are equiangular, 
Z apo= ahE; .-. opE = Fha= Fak; .«. 
op is parallel toaF. Which is Proposi- 29 
tion 29th. 
The 30th is very evident from the 30 
printed figure, for since AE . AF = AB? 
= AC . AD, the points C, D, E, F are in 
a circle, therefore the external angle ACE 
= EFD and = DHG;; .-. EFD being = 
and from A and B two lines be drawn to touch the circle 
in C and D, and let the square of AB be equal to the sum 
of the squares of AC and BD, and from A any line be 
drawn to meet the circle in E and F, and BE, BF he drawn 
meeting the circle again in G and H; the points A, G, H, 
will be in aright line. 
Prop. XXXIV. Let AB meet a circle in C and D, and 
A be without and B within the same, and let the rectangle 
CAD be equal to the square of AB together with the rect- 
angle CBD, and through A any line be drawn meeting the 
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