; .. AE = AG. Hence E and G coincide, and DE is || to BC Prop. 2, Bk. III., is also a demonstration of thistheorem. The preceding proof may be substituted forthat given on pages 112-3. .\dditional Peopositions. 235 The following is the usual figure and demonstrationof Prop. 13, Bk. III. The proof… … Let ABC be a right-angled A, having the rt. ZACB.Then the square described on AB is equal to the sumof the squares described on AC, CB. On AB, BC, CA describe the squares AE, BG, CK.Through C draw CM |i to AD or BE.Join CD, BK.Because the Zs ACH, ACB are right Zs, .. HC, CB are in the same st.…
Geometry · Geometry, Plane
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