Use Theorem 5.6.1 and the drawing to complete the proof

PERSPECTIVE ON HISTORY Ceva’s Proof Giovanni Ceva 1647–1736 was the Italian mathematician for whom Ceva’s Theorem is named. Although his theorem is difficult to believe, its proof is not lengthy. The proof follows. Proof Given with interior point D [see Figure 5.55a], draw a line through point C that is parallel to . Now extend to meet at point R . Likewise, extend to meet at point S. See Figure 5.55b. With similar triangles, we will be able to substi- tute desired ratios into the obvious statement , in which each numerator has a matching denominator. Because by AA, we have Also with , we have By the Transitive Property of Equality, and by interchanging the means, we see that [The first ratio, of this proportion will replace the ratio in the starred statement.] From the fact that , [The second ratio, of this proportion will replace the ratio in the starred statement.] With , [The first ratio, of this proportion replaces in the starred statement.] Making the indicated substitutions into the starred statement, we have AG GB BF FC CE EA = 1 CR AB CE EA , CE EA = CR AB . 䉭RCE 䉭BAE AB CS BF FC , AB SC = BF FC . 䉭CSF 䉭BAF CS CR AG BG , AG BG = CS CR . AG CS = GB CR , GD CD = GB CR . 䉭DGB 䉭DCR AG CS = GD CD . 䉭AGD 䉭SCD CS CR AB CS CR AB = 1 12 AF BE AB 䉭ABC PERSPECTIVE ON APPLICATION An Unusual Application of Similar Triangles The following problem is one that can be solved in many ways. If methods of calculus are applied, the solution is found through many complicated and tedious calculations. The simplest solution, which follows, utilizes geometry and similar triangles. Let point D be any point in the interior of , and let , , and be the line segments determined by D and vertices of . Then the product of the ratios of the segments of each of the three sides taken in order from a given vertex of the triangle equals 1; that is, AG GB BF FC CE EA = 1. 䉭ABC CG AF BE 䉭ABC THEOREM 5.6.4 왘 Ceva’s Theorem G a E F B A C D Figure 5.55 b G E F B A C D R S Problem: A hiker is at a location 450 ft downstream from his campsite. He is 200 ft away from the straight stream, and his tent is 100 ft away, as shown in Figure 5.56a on page 270. Across the flat field, he sees that a spark from his campfire has ignited the tent. Taking the empty bucket he is carrying, he runs to the river to get water and then on to the tent. To what point on the river should he run to minimize the distance he travels?