A treatise on surveying ; in which the theory and practice are fully explained (1860)
Alsop, Samuel · Philadelphia, E. C. & J. Biddle & co. · leaf 146
CK = CM, HL = HK, and AL == AK;also (47.1) BL = BM. Because AK = AL and BM = BL, CK -|- CM will be equal to the sum of the sides AB, AC, andBC; therefore CK or CM = £ (AB -f AC + BC) = J S, if S stand for thesum of the three sides. But CE -f AE -f- BG = \ S ; therefore CK = CM =CA + BG, and AK = AL… … we have (4.6)and (23.6) Whence, and CE : EDAE : EDAE . EC : ED2 v/AE.EC : ED : CK : KH,HK : KA,CK : KA :: CKa : CK . KA. CK : ^/CK . KA, CK . ED = vCK . KA . AE . EC. Now, ABC = ACD+ BCD + ABD == \ AC . ED+ \ BC . ED + \ AB . ED= JS . ED = CK. ED. Wherefore, ABC = y/CK . KA. AE . EC. Cor.—From the…
Surveying
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