CHAPTER 1 CEVA S THEOREM AND MENELAUS S THEOREM

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1 HTR 1 V S THOR N NLUS S THOR The purpose of this chapter is to develop a few results that may be used in later chapters. We will begin with a simple but useful theorem concerning the area ratio of two triangles with a common side. With this theorem in hand, we prove the famous eva s theorem and enelaus s theorem. The converses of these two theorems guarantee the existence of the centroid, incenter and orthocenter of any given triangle. s we will see in the examples, enelaus s theorem can be used to prove the Simson s theorem. ased on this, we will then go on to discuss the tolemy s theorem. These theorems are of the same importance. Notation. Given a triangle, we denote the length of three sides by a =, b =, c =. The length of three medians are denoted by m a, m b, m c, the length of three altitudes by h a, h b, h c, and the length of three angle bisectors by t a, t b, t c. The subscripts of these symbols indicate which median/altitude/angle bisector we are talking about. lso, the area of a triangle will be denoted by (). These are all standard notations used in many books. One more notation that is less standard: the semi-perimeter of a given triangle is usually denoted by p. In the case there is no risk of confusion, we will use these notations throughout this book without explanation anymore. 1.1 simple theorem on area ratio rea is one of the most intuitive concepts in mathematics. On one hand it is simple, people learn it since they were in primary school. On the other hand, it leads to an important notion called measure, which is a corner-stone of measure theory and even various branches of modern mathematics. In our situation, we are concerned in the techniques of using area to solve problems in geometry (especially those in Olympic level). Theorem 1.1-1( 共 邊 定 理 ) If the lines, Q intersect at, then ( ) =. ( Q) Q Theorem does not make any assumption on the positions of the points,,, Q. s we will see in the proof, there are four possible cases depending on the positions of these points. age 1 of 21

2 athematical atabase efore proving the theorem, let s recall that the area of a triangle is given by 1 ( ) = ah a. It means that if h a is fixed then the area is directly proportional to a. or example, 2 in igure 1 we have ( ) =. ( ) aking use of this observation we can have a short proof of Theorem igure 1 roof Without loss of generality we assume all triangles involved are not degenerated. Now, one has ( ) ( ) ( ) ( ) = Q ( Q) ( ) ( Q) ( Q) = Q = Q Q... Q Q Q Q igure 2 age 2 of 21

3 In order to familiarize ourselves with Theorem 1.1-1, we look at a few examples. athematical atabase xample Let be an interior point of triangle, the rays,, meet the sides,, at points,, respectively. rove that + + = 1. igure 3 Solution ( ) ( ) ( ) + + = + + ( ) ( ) ( ) ( ) = ( ) = 1 Q... xample (IO 1998 Hong Kong reliminary Selection ontest) In,,, G are points on,, respectively such that : = : = G : G = 1 : 3. K, L, are the intersection points of the lines and, G and, and G, respectively. Suppose the area of is 1; find the area of KL. K G L igure 4 age 3 of 21

4 Solution athematical atabase Let s = (L). y Theorem we have (L) = 3s and (L) = s / 3. Note that (L) + (L) + (L) = () = 1, s so we have s + + 3s = 1 and therefore 3 3 shows ( ) = ( K) =. Hence, 13 3 s =. We have proved that 13 ( KL ) = ( ) ( L) ( ) ( K) = = 13 3 ( L ) =. Similar argument 13 G K L igure 5 xample Refer to igure 6, there is a convex quadrilateral. The lines and intersect at K, the lines and intersect at L, the lines and KL intersect at G, the lines and KL intersect at. rove that K L KG =. GL K L G igure 6 age 4 of 21

5 Solution athematical atabase pply Theorem repeatedly, K L ( K) = ( L) ( K) ( KL) = ( KL) ( L) K = L ( ) ( K) = ( L) ( ) ( K) = ( L) KG = LG We will come back to this example later with two different proofs. One using eva s theorem and enulaus theorem (to be introduced in the next section), while another one involves the notion of cross ratio( 交 比 ) in projective geometry. xercise 1. Let be a triangle and, are points on the segment, respectively such that = λ and = µ. ind, in terms of λ and µ, the ratio :. igure 7 2. ( 定 比 分 點 公 式 )Suppose, Q are two points on the same side of the line. R is a point on the segment Q such that R = λq. rove that (R) = (1 λ)() + λ(q). 3. Refer to igure 8, is a convex quadrilateral. and intersect at., Q are the age 5 of 21

6 midpoints of and respectively. Given that = λ and = µ. (a) ind the ratios R : R and S : S (in terms of λ and µ). (b) Suppose the area of is 1. What is the area of SR? athematical atabase R Q S igure 8 4. Given a convex quadrilateral. Let 1, 2 be the trisection points of the segment and Q 1, Q 2 be the trisection points of the segment as shown in igure 9. rove that ( Q 1 2 2Q1) 1 ( ) = 3, where ( XYUV ) denotes the area of the quadrilateral XYUV. Q 1 Q igure 9 Refer to igure 10, we trisect, by the points R 1, R 2, S 1, S 2. rove that ( KLN) 1 ( ) = 9. age 6 of 21

7 athematical atabase S 2 N Q 1 Q 2 R 2 S 1 K L R igure eva s theorem, enelaus s theorem and their converses We are in a position to introduce two important theorems (and their converses) in elementary geometry, which are powerful tools for proving collinear points and concurrent lines. Theorem (eva s theorem) Let be a triangle and,, be points on the lines,, respectively. If,, are concurrent (i.e. meet at a point ), then =+ 1. The + sign emphasizes directed segments were used here. igure 11 roof The theorem can be proved easily by Theorem as follows: ( ) ( ) ( ) = = 1, ( ) ( ) ( ) age 7 of 21

8 and the sign is obviously positive. athematical atabase Q... Since Theorem doesn t depend on the positions of the points involved, the proof above is valid even for the case where lies outside the triangle. Theorem (onverse of eva s theorem) Let be a triangle and,, be points on the lines,, respectively. Suppose that =+ 1. Then,, are either concurrent or mutually parallel (sometimes we say the lines are concurrent at the point at infinity). The proof of Theorem is left to the reader as exercise. Theorem (enelaus s theorem) Let be a triangle and,, be points on the lines,, respectively. If,, are collinear, then = 1. X Y igure 12 roof Let X, Y be two arbitrary (distinct) points on the line. Then ( XY ) ( XY ) ( XY ) = = 1. ( XY ) ( XY ) ( XY ) age 8 of 21

9 gain, it is clear that the sign is negative in this case. athematical atabase Q... enelaus s theorem also has a converse: Theorem (onverse of enelaus s theorem) Let be a triangle and,, be points on the lines,, respectively. Suppose that = 1. Then,, are collinear. Here are some corollaries of the theorems above: Three medians of any given triangle are concurrent. The point of intersection is called the centroid( 重 心 )of the triangle. Three altitudes of any given triangle are concurrent. The point of intersection is called the orthocenter( 垂 心 )of the triangle. Three angle bisectors of any given triangle are concurrent. The point of intersection is called the incenter( 內 心 )of the triangle. oreover, the external bisectors of any two angles of a triangle are concurrent with the internal bisector of the third angle. The point of intersection is called an excenter( 旁 心 )of the triangle. Note that a triangle has three excenters. Usually, circumcenter, centroid, orthocenter and incenter are denoted by the letters O, G, H, I respectively. s we will see in chapter 2, for any given triangle the circumcenter O, the centroid G and the orthocenter H are collinear and OG : GH = 1 : 2. The line OGH is called the uler line( 歐 拉 線 )of the triangle. The following are some applications of eva s theorem, enelaus s theorem and their converses. Readers should be careful when applying these theorems we don t consider directed segments since the sign of an expression is usually obvious. age 9 of 21

10 xample (IO ) athematical atabase The diagonals and of the regular hexagon are divided by the inner points and N, respectively, so that / = N / = r. etermine r if,, N are collinear. Solution Join which intersects at. pply enelaus s theorem to the triangle and the line N, one has N (2.1) = 1. N N igure 13 Note that (i) (ii) (iii) 1 r 2 2r = = r 2r 1 ; = cos = = = ; N 1 r =. N r Substitute (i), (ii), (iii) into (2.1), 2 2r 1 1 r = 1 2r 1 4 r which implies 3 r =. 3 xample (lternative solution to xample 1.1-3) Refer to igure 14, there is a convex quadrilateral. The lines and intersect at K, the age 10 of 21

11 athematical atabase lines and intersect at L, the lines and KL intersect at G, the lines and KL intersect at. rove that K L KG =. GL K L G igure 14 Solution pply eva s theorem to triangle KL and the point, we have K L (2.2) = 1. K L pply enelaus s theorem to triangle KL and the line G, we have KG L (2.3) = 1. K GL ivide (2.2) by (2.3), the result follows. Q... efore looking at the third solution to this question, let s recall that the cross ratio of four (distinct) collinear points,,, is defined by {, } =. n ordered quadruple (,,, ) of four distinct collinear points is called a harmonic quadruple ( 調 和 四 元 組 )if {, } = 1. One may verify that {, } = {, } and {, } 1 =. {, } So, an ordered quadruple (,,, ) is harmonic if and only if {, } = {, }. age 11 of 21

12 Theorem (Invariant under perspectivity) athematical atabase Let L 1, L 2 be two distinct lines on the plane. If,,, are distinct points on L 1 and,,, are distinct points on L 2, and if the lines,,, are concurrent, then {, } = {, }. quivalently, =. Theorem says that cross ratio is an invariant under perspectivity( 中 心 投 影 ). O ' ' ' ' igure 15 roof Let,,, intersect at O and be the intersection of L 1, L 2 (when L1// L 2 we regard as the point at infinity, this proof is still valid). pply enelaus s theorem to the triangles,,, with the intersecting lines,,, respectively, ultiply these four equalities together, we obtain O = 1 O O = 1 O O = 1 O O = 1 O age 12 of 21

13 = 1. athematical atabase It follows that =. Q... Theorem has its origin in projective geometry which we will not pursue. We give an example to show how Theorem gives a beautiful solution of xample xample (The third solution to xample 1.1-3) Refer to igure 16, there is a convex quadrilateral. The lines and intersect at K, the lines and intersect at L, the lines and KL intersect at G, the lines and KL intersect at. rove that K L KG =. GL K L G igure 16 roof Let G, meet at. onsider the perspectivity of KG onto with center, by Theorem one has (2.4) { KL, G} = {, }. Next, we consider the perspectivity of onto L with center. y the same theorem one has (2.5) {, } = { LKG, }. Now, combining (2.4) and (2.5) gives { KL, G} = { LKG, }, age 13 of 21

14 saying that (K, L,, G) is a harmonic quadruple. athematical atabase Q... xercise 1. rove Theorem and Theorem (The 26 th and 31 st IO shortlisted problem) Let be an interior point of triangle. 1 meets at, meets at, meets at. rove that ( ) ( ) ( 張 角 公 式 )Suppose,, be three rays for which = + < 180. rove that,, are collinear if and only if sin sin sin = +. Using this result, find an alternative solution to xample (ascal s theorem) Let,,,,, be arbitrary (distinct) points on a given circle. rove that the intersections of with, with, and with are collinear if they exist. 5. (appus s theorem) If,, are three points on one line,,, on another, and if the three lines,, meet,,, respectively, then the three points of intersection L,, N are collinear. 6. (esargues s theorem) If two triangles are perspective from a point, and if their pairs of corresponding sides meet, then three points of intersection are collinear. 1.3 Simson s theorem and tolemy s theorem The next theorem, involving circles, possibly should not be put in this chapter. However, it is one of the famous applications of enelaus s theorem. age 14 of 21

15 athematical atabase Theorem (Simson s theorem) Let be a triangle. Suppose is a point on the circumcircle of triangle. Let,, be the feet of perpendicular from to,, respectively. Then,, are collinear. igure 17 roof To show,, are collinear, we need to verify = 1. Note that = cos, = cos, = cos, = cos, = cos, = cos. Therefore, cos cos cos =. cos cos cos Note also that =, =, =, the result follows. Q... The line is called the Simson line (or simply simson) of point with respect to triangle. The converse of Simson s theorem is also true. This is left to the reader as exercise. xample Refer to igure 18,,, are respectively the feet of perpendicular from to, to, and to. raw perpendicular lines from to,,, and let, Q,, N be the feet of perpendicular respectively. rove that, Q,, N are collinear. age 15 of 21

16 athematical atabase H N Q igure 18 Solution It is clear that H is a cyclic quadrilateral, the Simson line of passes,, N. In other words,,, N are collinear. Similar argument shows Q,, N are also collinear. Q... xample (IO 1998 shortlisted problem) Let be a triangle, H its orthocenter, O its circumcenter, and R its circumradius. Let be the reflection of across, be that of across, and that of across. rove that, and are collinear if and only if OH = 2R. O H igure 19 Solution Let QR be the triangle with as its medial triangle, i.e. is the midpoint of QR, is that of R and that of Q. rom O draw perpendicular lines to QR, R and Q with feet of perpendicular, and respectively. It can be proved, by considering a suitable homothety, that, and are collinear if and only if, and are collinear. We postpone the proof until chapter 5 in which we discuss geometric transformations. Readers who know homothety may try to prove it at this point. age 16 of 21

17 athematical atabase ' O Q ' R ' igure 20 y Simson s theorem (and its converse),, and are collinear if and only if O lies on the circumcircle of triangle QR. Note that the circumcenter of triangle QR is the orthocenter of triangle, namely H. So O lies on the circumcircle of triangle QR if and only if OH equals the circumradius of triangle QR, which is 2R. Q... Theorem (tolemy s theorem) or any four points,,, in general position (i.e. no two of them coincide, no three of them are collinear), +. quality holds if and only if is a cyclic quadrilateral. roof Let L,, N be respectively the feet of perpendicular from to,,. Since L = = 90, the points L,,, are concyclic. igure 21 shows one of the possible cases. In any case we have L = sin =, 2R where R denotes the circumradius of triangle. Similarly, we have N = and LN =. 2R 2R age 17 of 21

18 athematical atabase N L igure 21 y triangle inequality, L + N LN. Using the expressions of L, N, LN we obtained, it leads to +. 2R 2R 2R The required inequality is proved. quality holds if and only if L,, N are collinear, by Simson s theorem (and its converse) this happens if and only if lies on the circumcircle of triangle. Q... xample Let be an equilateral triangle and be a point on the circumcircle of triangle which lies on the arc. rove that = +. igure 22 Solution The result follows immediately by applying tolemy s theorem to the cyclic quadrilateral. Q... age 18 of 21

19 athematical atabase Here we are going to give an alternative solution to xample which uses only congruent triangles. Readers are encouraged to look at it carefully since it illustrates a standard technique which proves useful in problem involving broken segments. lternative Solution xtend to point such that =. Since = and = = 60, is an equilateral triangle. onsider triangles and. We have =, =, and = = 60 =. So, and hence = = +. igure 23 Q... xample If a circle passing through point cuts two sides and a diagonal of a parallelogram at points, Q, R as shown in igure 24, then + R = Q. R Q igure 24 Solution pply tolemy s theorem to the cyclic quadrilateral QR, we have (3.1) RQ + R Q = Q R. age 19 of 21

20 Observe that RQ. We multiply the constant / RQ to (3.1), it gives + R = Q. Replace by, we have + R = Q. athematical atabase Q... xample Let,,, be adjacent vertices of a regular 7-sided polygon, in that order. rove that = +. G igure 25 Solution Refer to igure 25, let,, G be the remaining vertices of the 7-sided polygon with the indicated order. pply tolemy s theorem to the cyclic quadrilateral : = +. Substitute by, by, by and by in the above equality, we obtain = +. ividing both sides by, the result follows. Q... xample ( Iranian ath Olympiad, IO 2000 Hong Kong Team Selection Test) is a triangle with > >. is a point on side, and is a point on produced beyond so that = =. Let be a point on side such that,,, are concyclic, and let Q be the second intersection point of with the circumcircle of. rove that Q + Q =. Solution We claim that Q ~. This is because Q = Q = and Q = 180 =. On the other hand, by tolemy s theorem, we have age 20 of 21

21 So = +. Q Q = + = + = Q + Q. athematical atabase Q... Q igure 26 xercise 1. State and prove the converse of Simson s theorem. 2. Suppose four lines intersect with each other and therefore any three lines among them determine a triangle. There are four such triangles. rove that the circumcircles of these triangles have a common point. 3. Let be a square. If is a point on the circumcircle of which lies on the arc, prove that the value ( + ) / does not depend on the position of. 4. Let be a regular pentagon inscribed in a circle O. is a point on O which lies on the arc. rove that + + = (IO 1995) Let be a convex hexagon with = =, = = and = = 60. Let G and H be two points in the interior of the hexagon such that G = H = 120. Show that G + G + GH + H + H. age 21 of 21

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