( 
X — 
TO GEOMETRICAL FORBLEMS. 
aa + 66 — cc , T 
265 
2 b 
: Now CD* = AC * 2 * — AD 4 = 
AC + AD x AC — AD = « + aa x 
26 
aa + bb — cc 2ab + aa + bb — cc 
a — —— x 
26 
2b 
a — b| 5 
2ab — aa — 66 + cc _ «+61* — c 1 c 1 
2b ~ <2b * 26 » 
hence CD = x \/ «+6? — c* x c* — 
and the area ( y ) — 
^/ a + 6? — c a x c z — a —T! 2 . 
But, because the difference of the squares of any two 
lines, or numbers, is equal to a rectangle under their 
sum and difference, the factor a •+- 6) 1 — c l will be — 
a + 6 + c X a + 6 — c ; and the remaining factor 
c* — a — b\ z — c -+ a — 6 X c — a + 6 ; and so 
the area will be likewise truly expressed by 
a + 6 c x « + 6 — c x c —J— « — 6 x c — « -+ b 
«t-6 + c a -+ 6 — c c + a — 6 c — a+ 6 
2 2 2 2 
« + 6 + C 
= V 
— \^s. s—c . s — 6.5 — a ; by making s ~ 
In order to determine the angles, which yet remain fo„ 
Be considered, we may proceed according to Prop. 11. 
in Trigonometry, by first finding the segments of the 
base : but there is another proportion frequently used 
in practice; which is thus derived: let BA be produced 
to F, so that AF may be =r AC ; and then I’C being 
joined, it is plain that the angle F will be the half of 
the angle A; and DF f= AC + AD) will be given