11. The following well-known argument illustrates the danger in relying too heavily on diagrams. Find the flaw in the “Proof." (The proof uses familiar high school notation that will be explained later in this textbook. For example, AB denotes the segment from A to B and AB denotes the line through points A and B.) False Proposition. If AABC is any triangle, then side AB is congruent to side AC. Proof. Let l be the line that bisects the angle ZBAC and let G be the point at which l intersects BC. Either l is perpendicular to BC or it is not. We give a different argument for each case. Assume, first, that l is perpendicular to BC (Figure 1.7). Then AAGB = AAGC by Angle- Side-Angle and therefore AB = AC. Now suppose l is not perpendicular to BC. Let m be the perpendicular bisector of BC and let M be the midpoint of BC. Then m is perpendicular to BC and l is not, so m is not equal to l and m is not parallel to l. Thus l and m must intersect at a point A D. Drop perpendiculars from D to the lines `AB and `AC and call the feet of those perpendiculars E and F, respectively. There are three possibilities for the location of D: either D is inside AABC, D is on AABC, or D is outside AABC. The three possibilities are B |G FIGURE 1.7: One possibility: the angle bi- illustrated in Figure 1.8. sector is perpendicular to the base Consider first the case in which D is on AABC. Then AADE = AADF by Angle-Angle-Side, so AE = AF and DE = DF. Also BD = CD since 3This fallacy is apparently due to W. W. Rouse Ball (1850–1925) and first appeared in the original 1892 edition of [3].