342 
THE CONSTRUCTION OP 
and therefore AD : FD : : AQ : HQ : : tang, of DHQ ; 
tang. D \Q. Likewise DA : DE(DH) :: sine of DHQ: 
sine of DAQ, as was to be shewn. 
METHOD OF CALCULATION. 
If AR be suppose 1 to meet the periphery io R, and 
R X 1 be drawn parallel to HF, meeting AK in N ; then 
will DA - DF, and A\ T : All : : AF : AH; but 
(by Euc. 37- 3.) AR : AK :: AE : AII ; whence, by 
compounding the terms of these two proportions, &c. 
AN: AK :: AF x Ah: AH'; whence AH, as Well 
as AD and DH, being known, the angles A and K will 
also be known. 
PROBLEM XXXII. 
To dr ate from a point A in the circumference of a gi 
ven circle, tuo subtenses AB and AD, which shall be to 
one another in the given ratio of m to n, and cut off two 
arches Ali and ABD, in the ratio of 1 to 3. 
CONSTRUCTION. 
Draw the diameter AH, 
and take the subtense AQ, 
in proportion thereto, as 
n — m to dm ; from the 
centre O draw OB paral 
lel to AQ, meeting the 
periphery in B ; join A, 
B.and make the subtenses 
BC and CD each equal 
to AB, and draw AD, and 
the thing is done. 
DEMONSTRATION. 
Join II, Q, and draw BE and CF perpendicular to 
AT). 
The angle AOB (QAII) at the ce ntre, s anding up 
on the arch AB, is equal to the angle BAD at the cir 
cumference, standing u, on double that arch; therefore, 
AQH being equal to ALB or a right angle (Euc. 31. 3.) 
thè triangles AQH, ALB, must be equiangular, and 
consequently AB : AE :: AH : ¿\Q ; but, by eon-