Triangles III Stewart s Theorem, Orthocenter, uler Line 23-Sept-2011 M 341 001 1 Stewart s Theorem (1746) With the measurements given in the triangle below, the following relationship holds: a 2 n + b 2 m = c(d 2 + mn) a d b m n c 24-Sept-2010 M 341 001 2 Stewart s Theorem (1746) so we will apply the Pythagorean Theorem several times a h d b p m n 24-Sept-2010 M 341 001 3 1
Stewart s Theorem (1746) In Δ a 2 = h 2 + (m p) 2 In Δ d 2 = h 2 + p 2 a 2 = d 2 -p 2 + (m p) 2 a 2 =d 2 + m 2 2mp a h d p m 24-Sept-2010 M 341 001 4 Stewart s Theorem (1746) In Δ b 2 = h 2 + (n + p) 2 b 2 = d 2 -p 2 + (n + p) 2 b 2 = d 2 + n 2 + 2np a h d p m 24-Sept-2010 M 341 001 5 Stewart s Theorem (1746) a 2 n =d 2 n+ m 2 n 2mnp b 2 m = d 2 m + n 2 m + 2mnp a 2 n + b 2 m = d 2 n + m 2 n + d 2 m + n 2 m = d 2 (n + m) + mn(m + n) a 2 n + b 2 m = c(d 2 + mn) 24-Sept-2010 M 341 001 6 2
The length of the median a m c m=c/2 n=c/2 24-Sept-2010 M 341 001 7 The length of the medians For a 3-4-5 triangle this gives us that the medians measure: 24-Sept-2010 M 341 001 8 xample 17 17 16 x x+8 Find x x=3 24-Sept-2010 M 341 001 9 3
Theorem 4 For any triangle, the sum of the lengths of the medians is less than the perimeter of the triangle. N in F so that NF=F N is a parallelogram N= In ΔN, N < +N 2F < + 2m a < b + c 24-Sept-2010 M 341 001 F N 10 Theorem 4 Similarly 2m b < a + c and 2m c < a + b 2(m a +m b +m c ) < 2a+2b+2c m a + m b + m c < a+ b+ c F 24-Sept-2010 M 341 001 N 11 Theorem 5 For any triangle, the sum of the lengths of the medians is greater than three-fourths the perimeter of the triangle. + > and F 24-Sept-2010 M 341 001 12 4
Theorem 5 24-Sept-2010 M 341 001 13 Result 24-Sept-2010 M 341 001 14 Theorem 6 The sum of the squares of the medians of a triangle equals three-fourths the sum of the squares of the sides of the triangle. 24-Sept-2010 M 341 001 15 5
Theorem 6 24-Sept-2010 M 341 001 16 Theorem 7 The sum of the squares of the lengths of the segments joining the centroid with the vertices is one-third the sum of the squares of the lengths of the sides. 24-Sept-2010 M 341 001 17 Theorem 8 median and the midline it intersects bisect each other. Show F and bisect each other. onstruct F and F. F F F 24-Sept-2010 M 341 001 18 6
Theorem 8 median and the midline it intersects bisect each other. F a parallelogram Thus, F and bisect each other. F 24-Sept-2010 M 341 001 19 Theorem 9 triangle and its medial triangle have the same centroid. This is HW Problem 2.1. F 24-Sept-2010 M 341 001 20 Orthocenter efinition: In Δ the foot of a vertex to the side opposite that vertex is called an altitude of the triangle. 27-Sept-2010 M 341 001 21 7
Orthocenter Theorem: The altitudes of a triangle meet in a single point, called the orthocenter, H. H We have shown this. 27-Sept-2010 M 341 001 22 uler Points The midpoint of the segment from the vertex to the orthocenter is called the uler point of Δ opposite the side. 27-Sept-2010 M 341 001 23 ircumcenter & entroid If the circumcenter and the centroid coincide, the triangle must be equilateral. Suppose =O, X = midpt of, Y=midpt of. =O =. =centroid 3/2 = 3/2 X=Y X=Y. y SS ΔY ΔX X=Y =. X Y 27-Sept-2010 M 341 001 24 8
The uler Segment The circumcenter O, the centroid, and the orthocenter H are collinear. Furthermore, lies between O and H and 27-Sept-2010 M 341 001 25 The uler Segment (Symmetric Triangles) xtend O twice its length to a point P, that is P = 2O. We need to show that P is the orthocenter. 27-Sept-2010 M 341 001 26 The uler Segment raw the median, L, where L is the midpoint of. Then, P = 2O and = 2L and by vertical angles we have that Then and OL is parallel to P. 27-Sept-2010 M 341 001 27 9
The uler Segment Since OL is perpendicular to, so it P, making P lie on the altitude from. Repeating this for each of the other vertices gives us our result. y construction P = 2O. This line segment is called the uler Segment of the triangle. 27-Sept-2010 M 341 001 28 Orthic Quadruple Let,,, and H be four distinct points with,, and noncollinear and H the orthocenter of Δ. Line determined by 2 of these points is perpendicular to the line determined by other 2 points!! 27-Sept-2010 M 341 001 29 Orthocenter of ΔH is the orthocenter of ΔH!!! 27-Sept-2010 M 341 001 30 10
Orthic Quadruple H is the orthocenter of Δ is the orthocenter of ΔH is the orthocenter of ΔH is the orthocenter of ΔH {,,, H} is called Orthic Quadruple 27-Sept-2010 M 341 001 31 Orthic Quadruple iven three points {,, } there is always a fourth point, H, making an orthic quadruple UNLSS 1.,, collinear 2.,, form a right triangle 27-Sept-2010 M 341 001 32 Orthic Triangle Let,, form a triangle and let,, F denote the intersections of the altitudes from,, and with the lines,, and respectively. The triangle F is called the orthic triangle. Theorem: The orthic triangles of each of the four triangles determined by an orthic quadruple are all the same. 27-Sept-2010 M 341 001 33 11
ircumcircle of Orthic Triangle onsider the circumcircle of the orthic triangle. It contains,, F of course!! 27-Sept-2010 M 341 001 34 ircumcircle of Orthic Triangle It also contains the uler points!! t least, it looks like it does. Prove it!! 27-Sept-2010 M 341 001 35 ircumcircle of Orthic Triangle It also contains the midpoints!! t least, it looks like it does. 27-Sept-2010 M 341 001 36 12
Nine Point ircle Theorem Theorem: For any triangle the following nine points all lie on the same circle: the three feet of the altitudes, the three uler points, and the three midpoints of the sides. Furthermore, the line segments joining an uler point to the midpoint of the opposite side is a diameter of this circle. Sometimes called Feuerbach s ircle. 27-Sept-2010 M 341 001 37 13