362 
THE CONSTRUCTION OU 
DEMONSTRATION. 
The triangles AEB and A^P, being similar, are to one 
another as the squares of their perpendicular heights FT 
and wîB («S); but »1B 1 is — BQ x AB rr gKF x AB ; 
therefore it wiil be, as the triangle AEB (EF x |AB) : 
the triangle A/iP :: EF J , : 2EF X AB :: EF : 2AB : : 
EF x -JAB : AB 1 (Eue. I. 6.) wherefore, the antece 
dents being the same, the consequents must necessarily 
be equal, that is, A/iP ABCD. Q. E. D. 
METHOD OF CALCULATION. 
In the triangle ABE are given all the angles and the 
side AB, whence EF will be given, and consequently 
S?i (= \/AB x 2EF); whence AP and An are also 
given. 
LEM MA. 
Jf from any point C, in one side of a plane angle K AL, 
a right line CB be drawn, cutting both sides AK,AL, in 
equal angles (ACB, ABC); and from any other point D 
in the same side AK another right line be draicn, to cut 
off an area ADE equal to the area ABC; I say y that 
DE will be greater than CB. 
DEMONSTRATION. 
Complete the parallelogram DCBG, and join B, D, 
and in BG (produced if need be) takeBF rz BE, and 
draw FD. 
Since the triangles ABC, AED are equal, by suppo 
sition, and have one 
. K. angle, A, common, 
•D/ therefore will AD : 
AC :: AB (AC) : 
AE (Euc. 15. 6.) 
and consequently 
AD p AE greater 
than AC 4- AB 
(Euc.25.5.) whence 
it is manifest that 
CD must be great 
er than EB, or BG 
than BE. Moreover, because the angle ABC (= ACB