Menelaus and Ceva theorems

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1 hapter 21 Menelaus and eva theorems 21.1 Menelaus theorem Theorem 21.1 (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.

2 332 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 It follows that collinear. = c b, = a c, = b a. = 1 and the points,, are

3 21.2 eva s theorem eva s theorem Theorem 21.2 (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.

4 334 Menelaus and eva theorems Example (1) The centroid. IfD, 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. F G E D 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.

5 21.2 eva s theorem 335 (3) In triangle, = 5π, = π, and = π. Prove that the altitude, the -bisector, and the -median are concurrent. Solution. T. Suppose =1. onsider the two squares P and P Q Note that triangle T is isosceles since T = T = π π = π =. Therefore, = , = = 2+ 2 = Since is the midpoint of, =1. The three lines,, are concurrent by eva s theorem. T

6 336 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 21.2 eva s theorem is a triangle with =12, =13, and =15. Show that the median D, angle bisector E, and the altitude F are concurrent. D E P F 4. Given three circles with centers,, and distinct radii, show that the exsimilicenters of the three pairs of circles are collinear.

8 338 Menelaus and eva theorems

9 hapter 22 Routh and eva theorems 22.1 arycentric coordinates In a given triangle, every point P is coordinatized by a triple of numbers (x : y : z) in such a way that the system of masses x at, y at, and z at will have its balance point at P. mass y at and a mass z at will balance at the point on the line. mass x at and a mass y + z at will balance at the point P. (y + z) = y + z, (x + y + z)p = x +(y + z) = x + y + z. We say that with reference to triangle, the point P has x + y + z (i) absolute barycentric coordinate and x + y + z (ii) homogeneous barycentric coordinates (x : y : z). (mass x) (mass x) P balance P balance (mass y) (mass z) (mass y + z)

10 340 Routh and eva theorems 22.2 evian and traces Let P be a point with homogeneous barycentric coordinates (x : y : z) with reference to triangle. The three lines joining a point P to the vertices of the reference triangle the cevians of P. The intersections,, of these cevians with the side lines are called the traces of P. The coordinates of the traces can be very easily written down: =(0:y : z), =(x :0:z), =(x : y :0). P Theorem 22.1 (eva theorem). Three points,, on,, respectively are the traces of a point if and only if they have coordinates of the form = 0 : y : z, = x : 0 : z, = x : y : 0, for some x, y, z.

11 22.2 evian and traces 341 Example The centroid. The midpoint points of the sides have coordinates = (0 : 1 : 1), = (1 : 0 : 1), = (1 : 1 : 0). The centroid G has coordinates (1 : 1 : 1). G Example The incenter I The traces of the incenter have coordinates = (0 : b : c), = (a :0:c), = (a : b :0). The incenter I has coordinates I =(a : b : c).

12 342 Routh and eva theorems 22.3 rea and barycentric coordinates Theorem If in homogeneous barycentric coordinates with reference to triangle, P =(x : y : z), then ΔP :ΔP :ΔP = x : y : z. (mass x) P balance (mass y + z) Proof. onsider the trace of P on. Since a mass x at and a mass y + z at balance at P, P : P = y + z : x, and P : = x : x + y + z. It follows that ΔP :Δ = P : = x : x + y + z. Similarly, ΔP :Δ = y : x + y + z and ΔP :Δ = z : x + y + z. ombining these, we have x : y : z =ΔP :ΔP :ΔP. ecause of this theorem, homogeneous barycentric coordinates are also known as areal coordinates.

13 22.3 rea and barycentric coordinates 343 useful area formula If for i =1, 2, 3, P i = x i + y i + z i (in absolute barycentric coordinates), then the area of the oriented triangle P 1 P 2 P 3 is x 1 y 1 z 1 ΔP 1 P 2 P 3 = x 2 y 2 z 2 x 3 y 3 z 3 Δ. Example Let,, be points on,, such that : =2:1, : =5:3, : =3:2. (The numbers indicated along the lines are proportions of lengths, and are not actual lengths) Their homogeneous barycentric coordinates are =0:1:2, = 3:0:5, and =2:3:0. Δ = Δ = Δ = 7 30 Δ.

14 344 Routh and eva theorems Example With the same points,, in the preceding example, the lines,, bound a triangle PQR. Suppose triangle has area Δ. Find the area of triangle PQR. 3 3 R 5 2 P Q 2 1 We have already known the coordinates of,,. From these, it is easy to find those of P, Q, R: P = Q = R = =(5:0:3) =(2:3:0) =(0:1:2) =(2:3:0) =(0:1:2) =(5:0:3) P = (10 : 15 : 6) Q =(2:3:6) R =(10:3:6) This means that the absolute barycentric coordinates of,, are P = ( ), Q = ( ), R = ( ) The area of triangle PQR = Δ= Δ.

15 22.3 rea and barycentric coordinates 345 Exercise 1. is a triangle of area 1, and 1, 2, 1, 2, 1, 2 are the points of trisection of, and respectively. Which of the two areas is larger, the one bounded by three lines 1, 1, 1 or the one bounded by the four lines 1, 2, 1, 2? Let,, be points dividing,, in the golden ratio. Let P be the intersection of and, Q that and, and R that of and. Show that (i) P, Q, R are respectively the midpoints of,, ; (ii) P divides Q in the golden ratio; so do Q and R divide R and P. (iii) ompare the areas of triangles and PQR. 1 1 rux Math. Problem nswer: 4.

16 346 Routh and eva theorems

Menelaus and Ceva theorems

Menelaus and Ceva theorems 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.