The Menelaus and Ceva Theorems

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1 hapter 7 The Menelaus and eva Theorems Sign convention Let and be two distinct points. point on the line is said to divide the segment in the ratio :, positive if is between and, and negative if is outside the segment. -1 < / < 0. / > 0. / < Harmonic conjugates Two points and Q on a line are said to divide the segment harmonically if they divide the segment in the same ratio, one externally and the other internally: = Q Q. We shall also say that and Q are harmonic conjugates with respect to the segment. 87

2 IU: Euclidean Geometry Let and Q be harmonic conjugates with respect to. If = d, = p, andq = q, thend is the harmonic mean of p and q, namely, 1 p + 1 q = 2 d. roof This follows from p d p = q d q. We shall use the abbreviation (, ;, Q) to stand for the statement, Q divide the segment harmonically. roposition If (, ;, Q), then (, ; Q, ), (, ;, Q), and (, Q;, ). Therefore, we can speak of two collinear (undirected) segments dividing each other harmonically. Exercise 1. Justify the following construction of harmonic conjugate. ' M Q Given, construct a right triangle with a right angle at and =. LetM be the midpoint of.

3 IU: Euclidean Geometry 89 For every point (except the midpoint of ), let 0 be the point on such that 0. The intersection Q of the lines 0 M and is the harmonic conjugate of with respect to. 7.2 pollonius ircle ngle bisector Theorem If the internal (repsectively external) bisector of angle intersect the line at (respectively 0 ), then : = c : b; 0 : 0 = c : b. c b c b ' : = c : b. : = c : -b Example The points and 0 areharmonicconjugateswithrespectto, since : = c : b, and 0 : 0 = c : b. and are two fixed points. For a given positive number k 6= 1, 1 the locus of points satisfying : = k :1isthecirclewithdiameter,where and are points on the line such that : = k :1 and : = k : 1. 1 If k 1, the locus is clearly the perpendicular bisector of the segment.

4 IU: Euclidean Geometry 90 roof. Since k 6= 1,points and can be found on the line satisfying the above conditions. onsider a point not on the line with : = k :1. Notethat and are respectively the internal and external bisectors of angle. This means that angle is a right angle. Exercise 1. The bisectors of the angles intersect the sides,, respectively at, Q, andr. 0, Q 0,andR 0 on the sides,, and respectivley such that 0 //, QQ 0 //, andrr 0 //. Show that QQ µ 1 RR 0 =2 a + 1 b + 1. c R Q' Q ' R' 2. Suppose is a triangle with 6=, andletd, E, F, G be points on the line defined as follows: D is the midpoint of, E is the bisector of 6, F is the foot of the perpeandicular from

5 IU: Euclidean Geometry 91 to, andg is perpendicular to E (i.e. G bisects one of the exterior angles at ). rove that = DF EG. G F E D k 3. If = d, andk 6= 1, the radius of the pollonius circle is k 2 1 d. 4. Given two disjoint circles () and(), find the locus of the point such that the angle between the pair of tangents from to () and that between the pair of tangents from to () areequal The Menelaus Theorem Let,, Z be points on the lines,, respectively. The points,, Z are collinear if and only if Z Z = 1. Z W Z W 2 Let a and b be the radii of the circles. Suppose each of these angles is 2θ. Then a =sinθ = b,and : = a : b. From this, it is clear that the locus of is the circle with the segment joining the centers of similitude of () and() asdiameter.

6 IU: Euclidean Geometry 92 roof. (= ) LetW be the point on such that W//. Then, = Z, and ZW = WZ Z. It follows that Z Z = Z ZW WZ Z Z Z = Z Z WZ ZW Z =( 1)( 1)( 1) = 1. Z ( =) Suppose the line joining and Z intersects at 0. From above, 0 0 Z = 1 = Z Z Z. It follows that 0 0 =. The points 0 and divide the segment in the same ratio. These must be the same point, and,, Z are collinear. Exercise 1. M is a point on the median D of 4 such that M : MD = p : q. The line M intersects the side at N. Find the ratio N : N. 3 2.Theincircleof4 touches the sides,, at D, E, F respectively. Suppose 6=. The line joining E and F meets at. Show that and D divide harmonically. F E F Z E D D 3 nswer: N : N = p :2q.

7 IU: Euclidean Geometry 93 3.Theincircleof4 touches the sides,, at D, E, F respectively. is a point inside 4 such that the incircle of 4 touches at D also, and touches and at and Z respectively. Show that E, F, Z, are concyclic. 4 I I ' ' 4. Given a triangle, let the incircle and the ex-circle on touch the side at and 0 respectively, and the line at and 0 respectively. Then the lines and 0 0 intersect on the bisector of angle, at the projection of on this bisector. 7.4 The eva Theorem Let,, Z be points on the lines,, respectively. The lines,, Z are concurrent if and only if Z Z =+1. roof. (= ) Supposethelines,, Z intersectatapoint.onsider the line cutting the sides of 4. y Menelaus theorem, 4 IMO = 1, or =+1.

8 IU: Euclidean Geometry 94 lso, consider the line Z cutting the sides of 4. y Menelaus theorem again, Z Z = 1, or Z Z =+1. Z Z Multiplying the two equations together, we have ( =) Exercise. 7.5 Examples The centroid Z Z =+1. If D, E, F are the midpoints of the sides,, of 4, then clearly F F D D E E =1. The medians D, E, F are therefore concurrent (at the centroid G of the triangle). onsider the line GE intersecting the sides of 4D. ythemenelau 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.

9 IU: Euclidean Geometry The incenter Let,, Z be points on,, such that then Z Z Z = b a, 6 bisects 6, 6 = c b, = a c. It follows that Z Z = b a c b a c =+1, and,, Z are concurrent, at the incenter I of the triangle. Exercise 1. Use the eva theorem to justify the existence of the excenters of a triangle. 2. Let,, Z be cevians of 4 intersecting at a point. (i) Show that if bisects angle and = Z, then 4 is isosceles. (ii) Show if if,, Z are bisectors and Z = Z, then 4 is a right triangle. 3. Suppose three cevians, each through a vertex of a triangle, trisect each other. Show that these are the medians of the triangle. 4. is a right triangle. Show that the lines, Q, andr are concurrent. Q R

10 IU: Euclidean Geometry If three equal cevians divide the sides of a triangle in the same ratio and the same sense, the triangle must be equilateral. 6. Suppose the bisector of angle, the median on the side b, andthe altitude on the side c are concurrent. Show that 6 cos α = c b + c. 7. Given triangle, construct points 0, 0, 0 such that 0, 0 and 0 are isosceles triangles satisfying 6 0 = 6 0 = α, 6 0 = 6 0 = β, 6 0 = 6 0 = γ. Show that 0, 0,and 0 are concurrent 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.Then = c b sin α 1. sin α 2 ~ ~ ~ ~ Z roof. ythesineformula, = / / = sin α 1/ sin β sin α 2 / sin γ = sin γ sin β sin α 1 = c sin α 2 b sin α 1. sin α 2 5 Klamkin 6 MME 263; MJ is the tangential triangle of.

11 IU: Euclidean Geometry Let,, Z be points on the lines,, respectively. The lines,, Z are concurrent if and only if roof. are nalogous to sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2 =+1. = a c sin β 1 sin β 2, = c b sin α 1 sin α 2 Multiplying the three equations together, Exercise Z Z = b a sin γ 1 sin γ 2. Z Z = sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ Show that the three altitudes of a triangle are concurrent (at the orthocenter H of the triangle). 2. Let 0, 0, 0 be points outside 4 such that 0, 0 and 0 are similar isosceles triangles. Show that 0, 0, 0 are concurrent. 8 I I I 8 Solution. Let be the intersection of 0 and. Then = sin(β+ω) sin(γ+ω) sin γ sin β.

12 IU: Euclidean Geometry Show that the perpendiculars from I to, fromi to, and from I to are concurrent Mixtilinear incircles Suppose the mixtilinear incircles in angles,, of triangle touch the circumcircle respectively at the points 0, 0, 0. The segments 0, 0,and 0 are concurrent. ' O K ' ' ~ ~ 2 ~ ~ Ñ ' roof. We examine how the mixtilinear incircle divides the minor arc of the circumcircle. Let 0 be the point of contact. Denote α 1 := 6 0 and α 2 := 6 0. Note that the circumcenter O, andthepointsk, 0 are collinear. In triangle KO, wehave OK = R ρ 1, O = R, 6 KO =2α 2, where R is the circumradius of triangle. Notethat 2 = b(s c) s,and K 2 = ρ pplying the cosine formula to triangle KO, wehave µ b(s c) 2 2R(R ρ 1 )cos2α 2 =(R ρ 1 ) 2 + R 2 ρ 2 1. s Since cos 2α 2 =1 2sin 2 α 2, we obtain, after rearrangement of the terms, sin α 2 = b(s c) s 9 onsider these as cevians of triangle I I I. 1 p 2R(R ρ1 ).

13 IU: Euclidean Geometry 99 Similarly, we obtain sin α 1 = c(s b) s 1 p 2R(R ρ). It follows that sin α 1 c(s b) = sin α 2 b(s c). If we denote by 0 and 0 the points of contact of the circumcircle with the mixtilinear incircles in angles and respectively, each of these divides the respective minor arcs into the ratios From these, sin β 1 a(s c) = sin β 2 c(s a), sin γ 1 b(s a) = sin γ 2 a(s b). sin α 1 sin β 1 sin γ 1 a(s c) b(s a) c(s b) = sin α 2 sin β 2 sin γ 2 c(s a) a(s b) b(s c) =+1. y the eva theorem, the segments 0, 0 and 0 are concurrent. Exercise 1. The mixtilinear incircle in angle of triangle touches its circumcircle at 0. Show that 0 is a common tangent of the mixtilinear incircles of angle in triangle 0 and of angle in triangle ' 10 roblem proposal to rux Mathematicorum.

14 IU: Euclidean Geometry Duality Given a triangle, let, 0, 0 be harmonic conjugates with respect to the side Z, Z 0. The points 0, 0, Z 0 are collinear if and only if the cevians,, Z Z are concurrent. Z' ' ' roof. y assumption, 0 0 =, 0 0 =, Z 0 Z 0 = Z Z. It follows that Z0 Z 0 = 1 if and only if Z Z =+1. The result now follows from the Menelaus and eva theorems Ruler construction of harmonic conjugate Given two points and, the harmonic conjugate of a point can be constructed as follows. hoose a point outside the line. Draw the

15 IU: Euclidean Geometry 101 lines,, and. Through draw a line intersecting at and at. LetZ be the intersection of the lines and. Finally, let Q be the intersection of Z with. Q istheharmonicconjugateof with respect to and Q O H H arm onic co njuga te ha rm o nic m ean Harmonic mean Let O,, be three collinear points such that O = a and O = b. IfH is the point on the same ray O such that h = OH is the harmonic mean of a and b, then(o, H;, ). Since this also means that (, ; O, H), the point H is the harmonic conjugate of O with respect to the segment. 7.9 Triangles in perspective Desargues Theorem Given two triangles and 0 0 0, the lines 0, 0, 0 are concurrent, 0 0 if and only if the intersections of the pairs of lines, 0 0 are, 0 0 collinear. roof. Suppose 0, 0, 0 intersect at a point. pplying Menelaus

16 IU: Euclidean Geometry 102 theorem to the triangle and transversal 0 0 R 0 0,wehave 0 0 Q 0 0 R R 0 0 = 1, = 1, 0 0 Q Q 0 0 = 1. Multiplying these three equation together, we obtain R R Q Q = 1. y Menelaus theorem again, the points, Q, R are concurrent. ' ' ' R Q Two triangles satisfying the conditions of the preceding theorem are said to be perspective. is the center of perspectivity, and the line QR the axis of perspectivity Given two triangles and 0 0 0, if the lines 0, 0, 0 are parallel, 0 0 then the intersections of the pairs of lines 0 0 are collinear. 0 0

17 IU: Euclidean Geometry 103 R ' ' Q ' roof. Q Q R R = µ 0 0 µ 0 0 µ 0 0 = If the correpsonding sides of two triangles are pairwise parallel, then the lines joining the corresponding vertices are concurrent. roof. Let be the intersection of 0 and 0.Then 0 = 0 0 = 0 0. The intersection of 0 and 0 therefore coincides with. ' ' ' Two triangles whose sides are parallel in pairs are said to be homothetic. The intersection of the lines joining the corresponding vertices is the homothetic center. Distances of corresponding points to the homothetic center are in the same ratio as the lengths of corresponding sides of the triangles.

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.