THE CONSTRUCTION OF 
PROBLEM LXX.HI« 
Having given the two opposite sides ah, cd, the two dia 
gonals ac, bd, and also the angle aeb in which they 
intersect each other; to describe the trapezium. 
CONSTRUCTION. 
In the indefinite line BP, take BD equal to bd, and 
make the angle DBF equal to the given angle aeb, and 
BF — ac; also from the centres D and F, with the 
radii dc and ah, let two arches mCn and rCs be de 
scribed intersecting each other in C ; join D, C, and F, 
C, and make BA equal and parallel to FC ; then draw 
AD, AC, and BC, and the thing is done. 
DEMONSTRATION. 
Since (by construction) AB is equal and parallel to 
CF, therefore will AC be equal and parallel to BF 
(Eue. 33. l.) and consequently the angle AEB (Eue. 2&. 
1.) — DBF r: aeb. Q. E. D. 
METHOD OF CALCULATION. 
Join D, F ; then in the triangle DBF wall be give* 
two sides DB, BF and the angle included, whence the 
angle BFD and the side DF will be know n ; then in 
the triangle DFC will be given all the three sides, 
whence the angle DFC will be known, from which 
BFC (BFD — DFC) — BAC will also be known. 
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