Menelaus and Ceva theorems

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1 hapter 3 Menelaus and eva theorems 3.1 Menelaus theorem Theorem 3.1 (Menelaus). Given a triangle with points,, on the side lines,, respectively, the points,, are collinear if and only if = 1. W Proof. (= ) Let W be the point on such that W//. Then, It follows that = W, and = W. = W W = W W = 1.

2 302 Menelaus and eva theorems ( =) Suppose the line joining and intersects at. From above, = 1 =. It follows that =. The points and divide the segment in the same ratio. These must be the same point, and,, are collinear. Example. The external angle bisectors of a triangle intersect their opposite sides at three collinear points. c b a Proof. If the external bisectors are,, with,, on,, respectively, then = c b, = a c, = b a. It follows that = 1 and the points,, are collinear.

3 3.1 Menelaus theorem 303 Exercise. 1. Given three circles with centers,, and distinct radii, show that the exsimilicenters of the three pairs of circles are collinear.

4 304 Menelaus and eva theorems 3.2 eva s theorem Theorem 3.2 (eva). Given a triangle with points,, on the side lines,, respectively, the lines,, are concurrent if and only if = +1. P Proof. (= ) Suppose the lines,, intersect at a point P. onsider the line P cutting the sides of triangle. y Menelaus theorem, P P = 1, or P P = +1. lso, consider the line P cutting the sides of triangle. y Menelaus theorem again, P P = 1, or P P = +1. Multiplying the two equations together, we have ( =) Exercise. = +1.

5 3.2 eva s theorem 305 Examples. (1) The centroid. If D, E, F are the midpoints of the sides,, of triangle, then clearly F F D D E E = 1. The medians D, E, F are therefore concurrent. Their intersection is the centroid G of the triangle. G onsider the line GE intersecting the sides of triangle D. y the Menelaus theorem, 1 = G GD D E E = G GD It follows that G : GD = 2 : 1. The centroid of a triangle divides each median in the ratio 2:1. (2) The incenter. Let,, be points on,, such that,, bisect angles, and respectively. Then = b a, = c b, = a c. I It follows that = b a c b a c = +1, and,, are concurrent. Their intersection is the incenter of the triangle.

6 306 Menelaus and eva theorems Exercise. 1. Given triangle with a = 15, b = 14, c = 9. (a) Find points on, on, and on such that = = and,, are concurrent. (b) Find also points on, on, and on such that = = and,, are concurrent. Q P 2. is a right triangle. Show that the lines,, and Q are concurrent. P Q

7 3.2 eva s theorem Let,, be cevians of intersecting at a point P. (i) Show that if bisects angle, and =, then is isosceles. (ii) Show if if,, are bisectors, and P P = P, then is a right triangle. 4. Suppose two cevians, each through a vertex of a triangle, trisect each other. Show that these are medians of the triangle. 5. is a triangle with = 15, = 13, and = 12. Show that the angle bisector D, the median E, and the altitude F are concurrent. D E P F 6. Suppose the bisector of angle, the median on the side b, and the altitude on the side c are concurrent. Show that 1 cosα = c b + c. 1 MME 263; MJ 455.

8 308 Menelaus and eva theorems Trigonmetric version of the eva Theorem Let be a point on the side of triangle such that the directed angles = α 1 and = α 2. y the sine formula, = / / = sin α 1/ sin β sin α 2 / sin γ = sin γ sin β sin α 1 sin α 2 = c b sin α 1 sin α 2. α 1 α 2 β 2 β 1 Likewise, if and be points on the lines, respectively, with = β 1, = β 2 and = γ 1, = γ 2, we have γ 2 γ 1 = a c sin β 1 sin β 2, = b a sin γ 1 sin γ 2. These lead to the following trigonometric version of the eva theorem. Theorem 3.3 (eva). The lines,, are concurrent if and only if sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2 = +1. Proof. = sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2.

9 3.2 eva s theorem 309 Exercise. 1. is an isosceles triangle with β = γ = 40. and are points on and respectively such that bisects angle and = 30. Let P be the intersection of and. Show that P =. P 2. Let be a triangle with = 40, = 60. Let E and F be points lying on the sides and respectively, such that E = 40 and F = 70. Let E and F intersect at P. Show that P is perpendicular to. F P E

10 310 Menelaus and eva theorems 3. In triangle, one pair of trisectors of the angles and meet at the orthocenter. (a) Show that the other pair of trisectors of these angles meet at the circumcenter. (b) Find all possible values of the angles of the triangle. O H H O H O

11 3.2 eva s theorem Let,,, D, E, F be six consecutive points on a circle. Show that the chords D, E, F are concurrent if and only if D EF = DE F. F E D is a regular 12-gon. Show that the diagonals 1 5, 3 6, and 4 8 are concurrent

12 312 Menelaus and eva theorems Excursus: Kiepert perspectors If similar isosceles triangles, and (of base angle θ) are constructed on the sides of triangle, either all externally or all internally, the lines,, and are concurrent. θ α 1 α 2 θ θ P θ β 2 β 1 θ θ γ 1 γ2 Proof. pplying the law of sines to triangles and, we have sin α 1 = sin α 1 sin(β + θ) sin α 2 sin(β + θ) sin(γ + θ) = sin(β + θ) sin(γ + θ) sin(β + θ) = sin(γ + θ). Likewise, sinβ 1 sinβ 2 = sin(γ+θ) sin(α+θ) and sinγ 1 sinγ 2 = sin(α+θ) sin(β+θ) sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2 = +1. and the lines,, are concurrent. sin(γ + θ) sin α 2. From these The point of intersection is called the Kiepert perspector K(θ). In particular, it is called (1) a Fermat point if θ = ±60, (2) a Napoleon point if θ = ±30, (3) a Vecten point if θ = ±45.

13 3.3 Harmonic conjugates Harmonic conjugates Two points and are said to divide two other points and harmonically if =. They are harmonic conjugates of each other with respect to the segment. Examples. 1. For two given circles, the two centers of similitude divide the centers harmonically. T T O I 2. Given triangle, let the internal bisector of angle intersect at. The harmonic conjugate of in is the intersection of with the external bisector of angle. c b 3. Let and be distinct points. If M is the midpoint of the segment, it is not possible to find a finite point N on the line so that M, N divide, harmonically. This is because N = M = 1, requiring N M N = N = N, and = N N = 0, a contradiction. We shall agree to say that if M and N divide, harmonically, then N is the infinite point of the line.

14 314 Menelaus and eva theorems onstruction of harmonic conjugate Given a point on a line, construct (1) triangle and the cevian line, (ii) points, on, respectively such that and intersect at a point on, (iii) the line to intersect, extended if necessary, at. Then is the harmonic conjugate of. Proposition 3.1. If and divide harmonically, then Proof. It follows from = that = 2. = =. Thus, ( ) = ( ), and (+) = 2. Dividing by, the result follows.

15 3.3 Harmonic conjugates 315 Exercise. 1. Justify the following construction of harmonic conjugate. M Given, construct a right triangle with a right angle at and =. Let M be the midpoint of. For every point (except the midpoint of ), let be the point on such that. The intersection of the lines M and is the harmonic conjugate of with respect to.

Advanced Euclidean Geometry

Advanced Euclidean Geometry dvanced Euclidean Geometry Paul iu Department of Mathematics Florida tlantic University Summer 2016 July 11 Menelaus and eva Theorems Menelaus theorem Theorem 0.1 (Menelaus). Given a triangle with points,,