37 6 
THE CONSTRUCTION OF 
This method of solving the problem, it may be ob 
served, requires three operations by the sines and 
tangents, but tire same thing may be performed by 
two proportions only: for as Er : AE :: the secant of 
rE>? to the tangent of E/iA ; whence all the rest will be 
found as above. 
PROBLEM LX IT. 
To reduce a given triangle into the form of another, or 
to make a triangle which shall be similar to one triangle, 
and equal to another, 
CONSTRUCTION. 
Upon the base AB of the triangle ABC, to which you 
would make another triangle equal, describe ADBsiini- 
U lar to the tri 
angle requi 
red; draw Cl 7 
parallel to AB 
meeting AD 
inF; take AE 
a mean pro 
portional be 
tween AD and 
ft AF ; and, pa 
rallel to DB, draw EG ; then will AGE be the triangle 
that was to be constructed. 
DEMONSTRATION. 
Let FR and DQ be perpendicular to AB; then the 
triang. ADB : triang. ACB :: DQ : FR (Schol. Euc s 
1. 6.) :: AD ; AF (Euc, 4. 6.) : : AD 1 : AD X AF 
[Ene. l. 6.) :: AD'“ : AE 1 [by construction) :: triang. 
ADB triang. AEG { Euc. 19. 6.). Therefore, the an 
tecedents of the first and last of these equal ratios being 
the same, the consequents ACB and AEG must neces 
sarily be equal. Q. E. D. 
METHOD OF CALCULATION. 
In the triangle ADB are given all the angles and the 
side AB, whence AD will be given; next, in the 
triangle AFR will be given all the angles and the side