EXERCISES 29 Consider triangles ABC, A B C such that AB = A B , AC = A C , and A A (Fig. 30). We want to prove that BC B C . Let us move the second triangle so that A B coincides with AB. Since A is less than A, side A C will move to a position AD in the interior of angle BAC. We construct the bisector AE of DAC. This segment is also inside BAC, and, as B and C lie on different sides of this line, it must intersect side BC in some point E located between B and C. If we draw DE, we see that triangles ACE and ADE are congruent, because they have an equal angle (with vertex A) between two corresponding equal sides (AE in common, AC = A C = AD). Therefore DE = EC. The inequality BD BE + ED provided by the triangle BDE then gives us BD BE + EC, or BD BC. QED Conversely. If, in two triangles, two pairs of sides are equal, but the third sides are unequal, then the angles opposite the unequal sides are unequal, and the greater angle is opposite the greater side. This statement is equivalent to the preceding one. Remark. The preceding theorem does not require the third case of congruence for triangles (SSS). It therefore provides another proof of that case. Indeed, if we have AB = A B , AC = A C and, in addition BC = B C , the angles A and A would have to be equal, or else BC and B C could not be equal. Knowing now that A = A the two triangles are congruent by the second case (SAS). Exercises Exercise 5. Prove that a triangle is isosceles: 1◦. if an angle bisector is also an altitude 2◦. if a median is also an altitude 3◦. if an angle bisector is also a median. Exercise 6. On side Ox of some angle, we take two lengths OA, OB, and on side Ox we take two lengths OA , OB , equal respectively to the first two lengths. We draw AB , BA , which cross each other. Show that point I, where these two segments intersect, lies on the bisector of the given angle. Exercise 7. If two sides of a triangle are unequal, then the median between these two sides makes the greater angle with the smaller side. (Imitate the con- struction in 25.) Exercise 8. If a point in the plane of a triangle is joined to the three vertices of a triangle, then the sum of these segments is greater than the semi-perimeter of the triangle if the point is inside the triangle, the sum is less than the whole perimeter. Exercise 8b. If a point in the plane of a polygon is joined to the vertices of the polygon, then the sum of these segments is greater than the semi-perimeter of the polygon.
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