435 
G E O IM E T R Y. 
173. (f loit/i the leajl fide AB of a given triang/e ABC, 
a fcmidrde be dtfcdbed from the angular point A j meeting the 
fide AC, produced in the points D, E ; and from B, the lines 
BE, BD, be drawn, and alfo BG perpendicular to DE: 
then the values of the feveral lines AG, CG, GE, GD, 
BE, BD, BG, may be exprejfed in terms of the fides of the 
triangle ABC, as follow. 
B. 
174. AGz= 
2 -2-s 
BC —AC —AB 
3 AC 
-2 -2 
Or /VG — . 
2AC 
175. CG— —^B -}-BC ^ For CG=AC + AG, 
2AC 
or to AC—AG. 
Ajid AC+AG=AC± 
BC —AC —AB 
2 AC 
-2 -2 —2 --2 
__2AC +BC —AC —AB 
2AC 
176. GEzz 2H=:AC+AB—BC. 
—— * T 77 =; ® 
For GE=AB(AE)±AG=:AB-1-^-. ^ 
2 ac 
__2 A C X AB + AC^+aF*—AC ^ 
~ 2AC 
But 2ACXAB4-AC +AB =(AC-i-AB =)CE , 
Then GEf —CE —BC \ CE+BCxC 
V 2 A c y 2 A c 
XCE—BC 
C A+AB+BC X C A-t-A B—BC 
2AC 
2HX2II—2BC 2HxH—BC 
2AC 
AC 
+ BC. 
For GD=ADq:AG-ABq:AG-AB-r ^^ 
2AC 
_2ACx AB—‘ac? —AB +BC* 
2AC 
. AC—AB +BC _ BC —CD' 
2AC 2AC 
._ BC—CDXBC + CD 
3 AC 
(“ 9 )- 
_ BC4-AB—ACxBC+AC—AB 
2AC 
. 2 H-2ACX2;i-2AB _ 
2AC 
178 
X2HXH—CB* 
For BE 2 =DExGE (170). 
-2 -2 
^ t' D 
rriABX-Therefore 
2AC 
BE=^/ —XCE^—BC^=J^X2HxH—CB- 
2AC ^ AC 
179. BD: 
AB 
AC 
XD—ACx 2H—2AC. 
For BD =:DExGD (170). 
=2ABx- 
B C —CD 
2 AC 
(iV?)” 
Tlterefore BD: 
2AB 
(177)- 
2ACXBC —CD 
180. BG=:^X^HxH—CBXH—AC xil—AB. 
For BG^=:GExGD. Therefore BG^v/GEXt/GD. 
And GE-~xHxH—CB. 
GDzr—XH—AC X 
H—AB. 
iSi. Carol. Hence.is derived the rule ufually giyen 
for finding the area, or fiiperficial content, of a triangle, 
the three fides being known.—ift. From half the fura 
of the three fides, fubtraft each fide feverally, noting 
the three remainders. 2d. Multiply the faid half fum, 
and the three noted remainders continually. 3d. The 
fquare root of the produdl is the area of the triangle. 
_ 182. Jf a regular polygon, ABC D E F, be infcribed in a 
circle-, and parallel to thefe fides if tangents 
drawn, meeting one another in the 
points a, b, c, d, e, f-, then fk,all 
the figure formed by thefe tangents 
circumfcribe the circle, arid be fimilar 
to the infcribed figure .—Since the 
circle touches every fide of the 
figure abcdef by confiruftion ; 
therefore the circle is circum- 
fcribed by that figure. Through 
A and B, draw the radii S A, 
SB, prolonged till they meet the tangent ab, in <2, b-,. 
then the triangles ASB, a'&b, are equiangular; for tlie 
^ at S is common ; and the other angles are equal,, be- 
caufe AB and ab are parallel, by i'uppofition. Alfo 
Sa = Si 5 ; for the triangles ASB, a^b, are ifofceles 
(104). And the fame may be proved of the other tri¬ 
angles ; and alfo, that they are equal to one another ; 
therefore the figure abcdef has equal- fides, and is equi¬ 
angular to the figure ABCDEF. Now SA : S« :: 
AB : ab-, and SA ; Sa :: AF : af (hC-;)-, therefore 
AB : ab :: Af' : af-, and the like of the other fides 
(<155),; confequently the figures ABCDEF, abcdef, 
are fimiiaiv( 145). 
183. Cor. I. If two figures are compofed oflike.fets 
of limilar triangles, thofe figures are fimilar. 
184. Cor. 2. Hence, if from the angles a, b, of a re. 
gular polygon circumfcribing a circle, lines aS, oS, be 
drawn to tl:e centre S ; the chord AB of the intercepted 
arc will be one fide. of. a fimilar polygon, infcribed in. 
the- 
