ecause CG = CA; .-. Z CGA = Z CAG.Hence the whole ZBGC = tlie whole ZBAC.That is, ZEDF = ZBAC.Evidently, in like manner, it follows that ZDEF =ZABC, and ZDFE = ZACB. That is, in the two As, the Zs which are oppositeequal sides are equal. The ZGBC being = ZABC, and ZGCB = ZACB, supposeA GBC to… … A; and the A GBC coincides with and is equal to theA ABC. That is, A DEF = A ABC iu area. Note 1. The pupil ^may be left to modifythe preceding proof tomeet the case whereAG falls without thetriangles, or passesthrough B or C. Note 2. Other waysof stating the preced-ing proj)Osition are tosay that…
Geometry · Geometry, Plane
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