GEOMETRICAL PROBLEMS 
2AB x CP - 2AH X / / / \ 
MN 4- MN 4 ; but / / / \ 
|AB x CP is - BH 4 / // V . 
— AIP (by a known / / I , , , X 
■property of triangles); I /A. P J) C 
therefore BH : — AH 4 / 
= £AH x M.V + MN* Ctl/ 
or BIl 4 — AH* 4 2 H x MN 4- MN 4 .AAH MN] X 
(Euc. 4. 2.) consequently BH = AH 4- MN. Q. B. D, 
METHOp OF CALCULATION. 
In the right-aqgled triangle A DE we have D E and 
whence the angle DAE 
(FEG) will be found; then in the triangle EFG will 
be given two sides and one angle, from which the angle 
Cl iv (— BAH) will also be known. 
PROBLEM LXXVII. 
The base, the perpendicular, and the sum. of the two 
sides being given, to describe the triangle. 
CONSTRUCTION. 
Bisect the base AB in C, and in it produced take CD 
a third proportional to sAB and the sum of the side's, 
MN; erect DE equal to the given perpendicular, and 
draw HE parallel to AB, and take therein EF MN; 
draw EAG, to which from F, apply FG n AB, draw 
AH parallel to FG, meeting EF in H, then draw BH, 
and the thing is done.