Triangles III. Stewart s Theorem (1746) Stewart s Theorem (1746) 9/26/2011. Stewart s Theorem, Orthocenter, Euler Line

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1 Triangles III Stewart s Theorem, Orthocenter, uler Line 23-Sept-2011 M 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 Stewart s Theorem (1746) so we will apply the Pythagorean Theorem several times a h d b p m n 24-Sept-2010 M

2 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 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 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

3 The length of the median a m c m=c/2 n=c/2 24-Sept-2010 M The length of the medians For a triangle this gives us that the medians measure: 24-Sept-2010 M xample x x+8 Find x x=3 24-Sept-2010 M

4 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 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 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

5 Theorem 5 24-Sept-2010 M Result 24-Sept-2010 M 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

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

6 Theorem 6 24-Sept-2010 M 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 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

7 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 Theorem 9 triangle and its medial triangle have the same centroid. This is HW Problem 2.1. F 24-Sept-2010 M 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

8 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 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 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

9 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 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 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

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.

10 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 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 Orthocenter of ΔH is the orthocenter of ΔH!!! 27-Sept-2010 M

11 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 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 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

12 ircumcircle of Orthic Triangle onsider the circumcircle of the orthic triangle. It contains,, F of course!! 27-Sept-2010 M ircumcircle of Orthic Triangle It also contains the uler points!! t least, it looks like it does. Prove it!! 27-Sept-2010 M ircumcircle of Orthic Triangle It also contains the midpoints!! t least, it looks like it does. 27-Sept-2010 M

13 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

Triangles III. Stewart s Theorem, Orthocenter, Euler Line. 23-Sept-2011 MA

Triangles III. Stewart s Theorem, Orthocenter, Euler Line. 23-Sept-2011 MA Triangles III Stewart s Theorem, Orthocenter, Euler Line 23-Sept-2011 MA 341 001 1 Stewart s Theorem (1746) With the measurements given in the triangle below, the following relationship holds: a 2 n +