1. Can a transversal intersect one side of a triangle internally and the other two sides externally?

Ans: No. Solution: Pasch's axiom says that if a line intersects one side of a triangle (internally), then it must intersect another side of the triangle (internally).

3. What does Menelaus' theorem say for triangle BCD and transversal AEF.

Solution:(BF/FD)(DE/EC)(CA/AB) = -1.

5. What does Menelaus' theorem say for triangle AEC and transversal FDB.

Solution: (AF/FE)(ED/DC)(CB/BA) = -1.

8. Prove that the internal bisectors of two angles of a triangle and the external bisector of the third angle intersect the opposite sides of the triangle in three collinear points.

Solution:

10.Altogether there are six bisectors of the angles of a triangle, three internal and three external. Prove that these six bisectors meet the opposite sides in six points that are on four lines, three on each line. (Compare this with a finite geometry studied in Chapter 1).

Solution:We know from exercise 8 that the points determined by two internal bisectors and one external bisector are collinear. This accounts for three lines having 3 points on each. Consider the three points determined by the three external bisectors.

Thus we have 4 lines through these 6 points. As every pair of the 6 points is on one of these 4 lines, there are no other lines passing through more than one of these 6 points. This set of points and lines forms a 4-line geometry.

13. Prove Ceva's theorem directly, without using the theorem of Menelaus.

Solution:

15. Use Theorem 4.13 to prove that the internal bisectors of the angles of a triangle are concurrent.

Solution: This is similar to exercise 8. Because the third bisector is now internal instead of external, the ratio it is involved in is positive instead of negative. The product now equals +1, and so, by Ceva's theorem, the three internal bisectors are concurrent.

17. How many Simson lines are associated with one triangle?

Ans:Infinitely many. Solution: There are an infinite number of points on the circumcircle from which to drop perpendiculars. A Simson line can only come from one such point, since once the three points of intersection with the triangle sides are determined, there can be only one point of intersection of the perpendiculars erected at those points.