hapter 1 Theorems of eva and Menelaus We start these lectures by proving some of the most basic theorems in the geometry of a planar triangle. Let,, be the vertices of the triangle and,, be any points on the lines, and respectively. For arbitrarily chosen points,, there is no reason to expect that the three segments,, will meet at one point. However they do meet in many cases of geometrical significance, e.g. the three angle bisectors meet at one point. efore we discuss these special cases, we will formulate a general criterion that tells us when the segments,, are concurrent. This criterion is called eva s theorem. Theorem 1 (eva). Let be a triangle and,, be points on the sides, and respectively. Three lines,, are concurrent if and only if = 1 We will prove this theorem using ratios of the areas of the triangles created by the segments, and. 1
2 HTER 1. LETURE We will rely on the following lemma: Lemma 2. Let be an arbitrary triangle and let D be a point on the side. Then rea(d) = D rea(d) D roof. Triangles D and D share a common height from the vertex. Let h be the length of this height. Then rea(d) = 1h D, 2 rea(d) = 1 rea(d) h D and hence = D. 2 rea(d) D We will in fact use the following corollary from this lemma. orollary 3. Let be a triangle with a point D lying on the side and point any point on the segment D. Then rea( ) rea( ) = D D D roof. From lemma 2 applied to triangle we get rea(d) = D D rea(d) From the same lemma applied to triangle we get rea(d ) = D D rea(d )
3 y subtracting these two from each other we get as needed. rea( ) = D D We are ready now to prove eva s theorem. rea( ) roof. For one of the implications suppose that the segments,, meet at a point. From corollary 3 we find that = rea( ) rea( ) = rea( ) rea( ) = rea( ) rea( ) Multiplying these equations we get that = 1, as needed. For the other implication, assume that points,, on the sides,, satisfy = 1. Let be the point of intersection of and. Extend the segment to meet at. Then, by the direction we have already proved, ut we assumed also that = 1 = 1 Hence =, meaning that and are in fact the same point. Thus the point lies on the segment as well. Remark 1. eva s theorem has analogues both in spherical and hyperbolic geometries. Instead of quotients of lengths of sides, quotients of certain functions of these lengths appear there (sines for spherical geometry and hyperbolic sines for hyperbolic geometry). The proof we supplied can be adapted to these geometries as well, except other formulas for areas of triangles will be needed.
4 HTER 1. LETURE We can also find the ratio in which the point divides the cevians, and in terms of the ratios in which the points, and divide the sides of the triangle. Indeed, it follows from the lemma that rea( ) = rea( ) and similarly that rea( ) = rea( ). y adding these two equalities we get that rea( ) + rea( ), or corrollary we know that = rea( ) = rea( ) rea( ) rea( ) = + + rea( ) rea( ) rea( ) =. ut from the and rea( ) rea( ) =. Thus We will now apply eva s theorem to prove a couple of classical results. Recall that a median in a triangle is a line connecting a vertex with the midpoint of the opposite side. n angle bisector is a line dividing an internal angle of a triangle at a vertex into two equal parts. n altitude is a perpendicular dropped from a vertex to the opposite side of the triangle. We will prove that the three medians of any triangle are concurrent. The same holds for angle bisectors and altitudes. We will first derive these results from eva s theorem and then rederive them in an independent way. Medians If,, are midpoints of the sides of the triangle, then = 1 1 1 = 1, and hence, by eva s theorem the three medians, and are concurrent. Let s prove the same result more directly. We claim that the median is exactly the locus of points in the plane such that rea( ) = rea( ). Indeed, if the point D is the intersection point of and, then the quotient rea( ) D is equal to rea( ) D by corollary 3. Hence it can be equal to 1 if and only if the point D coincides with the midpoint of ; that is lies on. Now let be the point of intersection of medians and. y the result we just showed rea( ) = rea( ) (because is on ) and rea( ) = rea( ) (because is on ). ombining these two equalities we get that rea( ) = rea( ), hence the point lies on the median as well. ngle bisectors To prove that three angle bisectors meet at one point, we will use the following lemma: Lemma 4. Let be a point on the side so that is the internal angle bisector of the angle at vertex in triangle. Then =.
roof. From lemma 2 we know that. Now the equality of rea( ) angles and implies that if we reflect the triangle in the line, then the reflected copy will share an angle with triangle and the reflection of will lie on the line (see pic.). = rea( ) 5 ut in the picture after the reflection we see that rea( ) rea( ) =. This lemma implies that if, and are internal bisectors of triangle, then = = 1, so eva s theorem tells us that, and are concurrent. We can prove it in a different way as well. For this we first notice that the internal angle bisector of an angle with vertex is exactly the locus of points inside the angle that are equidistant from the two rays that form the angle. Indeed, let and be the feet of perpendiculars dropped from the point to the two rays. If the point is on the angle bisector, then the reflection of triangle in line must coincide with the triangle (because = and = π/2 = π/2 = ), and hence =. onversely, if is a point inside the angle with =, then in the right triangles and side is common, and sides and are equal. ut since there is only one right triangle with a given leg and hypothenuse, the triangles must be equal. In particular this implies that =, so the point lies on the angle bisector of angle. Now let, and be the internal angle bisectors of triangle. Let be the point of intersection of the angle bisectors and.
6 HTER 1. LETURE We have proved above that point is equidistant from sides and because it lies on angle bisector. It is also equidistant from sides and, because it lies on the angle bisector. Hence the point is located at the same distance from the sides and. It must then lie on the angle bisector. Note that since the point of intersection of the angle bisectors of triangle is equidistant from the sides of the triangle, it must be the center of the circle inscribed in triangle. s another corollary from eva s theorem, we can prove that the three altitudes of any triangle are concurrent (for obtuse triangles the point of concurrency lies outside the triangle, so some care must be taken in this case). Indeed, if and are altitudes in a triangle, then the triangles and are similar, since they share an angle at and are both right-angled. Hence =. pplying this argument three times we get that the feet of the altitudes satisfy = = = 1. eva s theorem then implies that the three altitudes are concurrent. Now we will reprove this result without reference to eva s theorem. Instead we will apply a technique called angle chasing repeatedly noting relations between different angles until we get what we need. Let and be altitudes of triangle and let point be their point of intersection. onnect points and by a line and let be the point of intersection of this line with side. We want to show that angle
is right. For this note that = = π/2, so both points and lie on the circle having as a diameter. The angles and are supported by the same arc in this circle, and therefore they are equal. Similarly, since both angles and are right, the points and lie on the circle having as a diameter. ngles and are supported by the same arc in this circle, and hence are equal. ombining these two results we get that = = = =. This means that in triangles and the angle at is common, and =, so the other two angles must be equal as well: = = π/2, as required. 7 *dd exercise proving the theorem from the corresponding theorem about side bisectors. There is a theorem, which is very close to eva s theorem in spirit and in formulation, called Menelaus theorem. Let s see what it states: Theorem 5. Let be a triangle and let points, and belong to the lines, and respectively. The points, and belong to one line if and only if = 1 The reader may object that there must be some mistake in this formulation, because the condition = 1 from Menelaus theorem is exactly the same as the condition = 1 that appeared in eva s theorem. If both theorems are correct, which one applies to any given triangle? One way to distinguish the cases in which the two theorems apply is to count the number of points, and that lie outside of the corresponding sides, and - if none or two, then eva s theorem applies; if one
8 HTER 1. LETURE or three - then Menelaus applies. It is, however, much more pleasing (both aesthetically and for generalizing both theorems) to treat both formulations on common grounds. For this we introduce the notion of oriented lengths. To speak of oriented length we need a directed line (that is a line with a choice of one of the two possible directions of movement along it) and a directed segment on it (that is a segment for which we can distinguish its beginning and its end). The oriented length of a segment from point, on a directed line, to point on the same line, is just its usual length, if the direction from to agrees with the chosen direction along the line; otherwise, it is the usual length taken with sign minus. The oriented length of the segment from to on the oriented line from figure 1 is positive, while the oriented length of the segment from to is negative. Notice that if we change the orientation of the line to the opposite one, the signs of the lengths will change as well. For a segment on a non-oriented line the sign of the length doesn t have any meaning (there is no reason to prefer moving in one of the directions along a line to moving in the other). However when we have two segments on the same line, the quotient of the oriented lengths makes sense: we can choose either direction along the line and compute the quotient of oriented lengths. If we choose the other direction instead, both oriented lengths will change sign, but their quotient won t change. D Figure 1: While the oriented lengths of segments and D don t have any meaning without the choice of direction along the line containing them, their quotient is well defined. In fact = = = D D D D D Now, in the formulations of eva s and Menelaus theorems the conditions = 1 and = 1 should be regarded as conditions on quotients of oriented lengths. When considered this way, they are indeed different: for eva s theorem the condition is = 1 while for Menelaus theorem it is = 1; Question 1. Why does the condition = 1 of eva s theorem imply that either none or two of the points, and lie outside the
corresponding segments, and? How many of them should lie outside the segments, and if the condition of Menelaus theorem applies? Having clarified the formulations of eva s and Menelaus theorems, we proceed to proving the latter. The following lemma will be helpful to us: Lemma 6. Let l 1, l 2 be two lines. Let points 1, 1, 1, D 1 belong to line l 1 and points 2, 2, 2, D 2 - to line l 2. Suppose that the lines 1 2, 1 2, 1 2 and D 1 D 2 are parallel. Then 1 1 1 D 1 = 2 2 2 D 2. This lemma can be interpreted as saying that the quotient of oriented lengths of segments is preserved under parallel projection from one line to another. roof. onsider the case when lines l 1, l 2 are parallel first. If we translate the line l 1 so that the point 1 gets translated to point 2, the points 1, 1 and D 1 get translated to points 2, 2 and D 2. Of course, translation preserves not just the quotients of lengths, but the lengths themselves. Suppose then that the lines l 1 and l 2 intersect at a point O. In that case considerations of similar triangles show that there is a number k so that k = O 1 O 2 = O 1 O 2 = O 1 O 2 = OD 1 OD 2. ut then 1 1 as we wanted to show. 1 D 1 = O 1 O 1 OD 1 O 1 = k(o 2 O 2 ) = 2 2 k(od 2 O 2 ) 2 D 2 9 Now we can prove the theorem of Menelaus. roof. Suppose that the points,, lie on one line. hoose any line l which is not parallel to it, and consider a parallel projection of all the picture to the line l along the line. Then the points,, project to the same point O on the line l. Denote by l, l, l the images of points,, under this projection. Then the lemma implies that O l O l, = O l O l, = O l = O l O l O l O l O l O l = O l. y multiplying these three equalities we get that = 1 (because of cancellation). The proof of the converse direction is similar to the proof of the converse direction in eva s theorem: suppose = 1 and let be the point on line at which the line intersects it. Then what we proved shows that = 1 and hence =, showing that = ; that is the point lies on the line.
10 HTER 1. LETURE l l l l 1.1 The relationship between eva s and Menelaus theorems One may wonder whether the results of eva and Menelaus, which are so similar in nature and formulation, can be deduced from each other. nd indeed it is the case. Suppose we know Menelaus theorem and want to deduce eva s: let be a triangle and let cevians, and meet at a point. Then, the application of Menelaus theorem to triangle and line gives that = 1. The application of Menelaus theorem to triangle and line gives = 1. y multiplying these equations we get = 1. Notice the ratios coming from the first product and coming from the second product are inverses of each other. Notice also that the side
1.1. THE RELTIONSHI ETWEEN EV S ND MENELUS THEOREMS11 appears in the numerator in the first product and in the denominator in the second one. Thus cancelling like terms we find = 1. This proves eva s theorem. For the derivation of Menelaus theorem from the theorem of eva we refer the reader to exercise 1 Exercise 1. In this exercise we are going to deduce Menelaus theorem from corollary of eva s theorem. Let be a triangle and let,, be three collinear points on lines,, respectively. Let be the point of intersection of lines and. 1) In triangle the evians,, are concurrent. Verify that the following equalities follow from eva s theorem and its corollary: = 1 = + 2) Deduce from the two equations above that or = 1. = (1 + ) =