A Special Point on the Median
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Point of Concurrency the Three Perpendicular Bisectors of a Triangle Intersect at a Single Point
3.1 Special Segments and Centers of Classifications of Triangles: Triangles By Side: 1. Equilateral: A triangle with three I CAN... congruent sides. Define and recognize perpendicular bisectors, angle bisectors, medians, 2. Isosceles: A triangle with at least and altitudes. two congruent sides. 3. Scalene: A triangle with three sides Define and recognize points of having different lengths. (no sides are concurrency. congruent) Jul 249:36 AM Jul 249:36 AM Classifications of Triangles: Special Segments and Centers in Triangles By angle A Perpendicular Bisector is a segment or line 1. Acute: A triangle with three acute that passes through the midpoint of a side and angles. is perpendicular to that side. 2. Obtuse: A triangle with one obtuse angle. 3. Right: A triangle with one right angle 4. Equiangular: A triangle with three congruent angles Jul 249:36 AM Jul 249:36 AM Point of Concurrency The three perpendicular bisectors of a triangle intersect at a single point. Two lines intersect at a point. The point of When three or more lines intersect at the concurrency of the same point, it is called a "Point of perpendicular Concurrency." bisectors is called the circumcenter. Jul 249:36 AM Jul 249:36 AM 1 Circumcenter Properties An angle bisector is a segment that divides 1. The circumcenter is an angle into two congruent angles. the center of the circumscribed circle. BD is an angle bisector. 2. The circumcenter is equidistant to each of the triangles vertices. m∠ABD= m∠DBC Jul 249:36 AM Jul 249:36 AM The three angle bisectors of a triangle Incenter properties intersect at a single point. -
Median and Altitude of a Triangle Goal: • to Use Properties of the Medians
Median and Altitude of a Triangle Goal: • To use properties of the medians of a triangle. • To use properties of the altitudes of a triangle. Median of a Triangle Median of a Triangle – a segment whose endpoints are the vertex of a triangle and the midpoint of the opposite side. Vertex Median Median of an Obtuse Triangle A D Point of concurrency “P” or centroid E P C B F Medians of a Triangle The medians of a triangle intersect at a point that is two-thirds of the distance from each vertex to the midpoint of the opposite side. A D If P is the centroid of ABC, then AP=2 AF E 3 P C CP=22CE and BP= BD B F 33 Example - Medians of a Triangle P is the centroid of ABC. PF 5 Find AF and AP A D E P C B F 5 Median of an Acute Triangle A Point of concurrency “P” or centroid E D P C B F Median of a Right Triangle A F Point of concurrency “P” or centroid E P C B D The three medians of an obtuse, acute, and a right triangle always meet inside the triangle. Altitude of a Triangle Altitude of a triangle – the perpendicular segment from the vertex to the opposite side or to the line that contains the opposite side A altitude C B Altitude of an Acute Triangle A Point of concurrency “P” or orthocenter P C B The point of concurrency called the orthocenter lies inside the triangle. Altitude of a Right Triangle The two legs are the altitudes A The point of concurrency called the orthocenter lies on the triangle. -
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10.4 - Circles, Arcs, Inscribed Angles, Power of a Point We work in Euclidean Geometry for the remainder of these notes. Definition: A minor arc is the intersection of a circle with a central angle and its interior. A semicircle is the intersection of a circle with a closed half-plane whose boundary line passes through its center. A major arc is the intersection of a circle with a central angle and its exterior (that is, the complement of a minor arc, plus its endpoints). Notation: If the endpoints of an arc are A and B, and C is any other point of the arc (which must be used in order to uniquely identify the arc) then we denote such an arc by qACB . Measures of Arcs: We define the measure of an arc as follows: Minor Arc Semicircle Major Arc mqACB = µ(pAOB) mqACB = 180 m=qACB 360 - µ(pAOB) A A A C O O C O C B B B q q Theorem (Additivity of Arc Measure): Suppose arcs A1 = APB and A2 =BQC are any two arcs of a circle with center O having just one point B in common, and such that their union A1 c q A2 = ABC is also an arc. Then m(A1 c A2) = mA1 + mA2. The proof is straightforward and tedious, falling into cases depending on whether qABC is a minor arc or a semicircle, or a major arc. It is mainly about algebra and offers no new insights. We’ll leave it to others who have boundless energy and time, and who can wring no more entertainment out of “24.” Lemma: If pABC is an inscribed angle of a circle O and the center of the circle lies on one of its 1 sides, then μ()∠=ABC mACp . -
Exploring Some Concurrencies in the Tetrahedron
Exploring some concurrencies in the tetrahedron Consider the corresponding analogues of perpendicular bisectors, angle bisectors, medians and altitudes of a triangle for a tetrahedron, and investigate which of these are also concurrent for a tetrahedron. If true, prove it; if not, provide a counter-example. Let us start with following useful assumption (stated without proof) for 3D: Three (non-parallel) planes meet in a point. (It is easy to see this in a rectangular room where two walls and the ceiling meet perpendicularly in a point.) Definition 1: The perpendicular edge bisector is the A generalisation to 3D of the concept of a perpendicular bisector. For example, the perpendicular edge bisector of AB is the plane through the midpoint of AB, and perpendicular D to it. Or equivalently, it is the set of points equidistant from A and B. B Theorem 1: The six perpendicular edge bisectors of a C tetrahedron all meet at the circumcentre. Proof: Let S be the intersection of the three perpendicular edge bisectors of AB, BC, CD. Therefore SA = SB = SC = SD. Therefore S must lie on all 6 perpendicular edge bisectors, and the sphere, centre S and radius SA, goes through all four vertices. Definition 2: Let a, b, c, d denote the four faces of a A tetrahedron ABCD. The face angle bisector of ab is the plane through CD bisecting the angle between b the faces a, b. It is therefore also the set of points equidistant from a and b. D a Theorem 2: The six face angle bisectors of a B tetrahedron all meet at the incentre. -
Power of a Point Yufei Zhao
Trinity Training 2011 Power of a Point Yufei Zhao Power of a Point Yufei Zhao Trinity College, Cambridge [email protected] April 2011 Power of a point is a frequently used tool in Olympiad geometry. Theorem 1 (Power of a point). Let Γ be a circle, and P a point. Let a line through P meet Γ at points A and B, and let another line through P meet Γ at points C and D. Then PA · PB = PC · P D: A B C A P B D P D C Figure 1: Power of a point. Proof. There are two configurations to consider, depending on whether P lies inside the circle or outside the circle. In the case when P lies inside the circle, as the left diagram in Figure 1, we have \P AD = \PCB and \AP D = \CPB, so that triangles P AD and PCB are similar PA PC (note the order of the vertices). Hence PD = PB . Rearranging we get PA · PB = PC · PD. When P lies outside the circle, as in the right diagram in Figure 1, we have \P AD = \PCB and \AP D = \CPB, so again triangles P AD and PCB are similar. We get the same result in this case. As a special case, when P lies outside the circle, and PC is a tangent, as in Figure 2, we have PA · PB = PC2: B A P C Figure 2: PA · PB = PC2 The theorem has a useful converse for proving that four points are concyclic. 1 Trinity Training 2011 Power of a Point Yufei Zhao Theorem 2 (Converse to power of a point). -
Group 3: Clint Chan Karen Duncan Jake Jarvis Matt Johnson “The
Group 3: Clint Chan Karen Duncan Jake Jarvis Matt Johnson “The power of a point P with respect to a circle is the product of the two distances PA times PB from the point to the circle measured along a random secant” (B&B 261). When the point P is on the circle, its power is zero; when it is inside the circle, its power is negative; and when it is outside the circle, its power is positive. The power of a point when P is outside the circle is also equal to (PT)^2, where T is the point where the line through P is tangent to the circle. Also of interest, if P is outside the circle and e is a circle which is orthogonal to c and centered at P, then pc(A) = t^2, where t is the radius of e (from "The Power of a Point and Radical Axis" class notes). The radical axis of two circles is, by definition, the locus of points A for which the power of A with respect to c[1] and c[2] are equal. Using some facts about the power of a point, we can see that the radical axis is a line. Fact 1: If A is a point outside the circle and S is a secant intersecting c at M and N, then the power of A with respect to c is equal to _AM__AN_. Fact 2: If A is a point outside the circle and AT s a line tangent to c at T, then the power of A with respect to c is equal to _AT_^2. -
Midpoints and Medians in Triangles
MPM1D Midpoints and Medians in Triangles Midpoint: B A C 1) Use a ruler to create a midpoint for AB. Label it X. 2) Use a ruler to create a midpoint for AC. Label it Y. 3) Join the points X and Y with a line. 4) Measure all the sides. Write them on the diagram. Q: What do you notice about the lenGth of XY and BC? Q: What do you notice about the direction of the lines XY and BC? Q: How could we prove this? The line seGments joining the midpoint of two sides of a triangle is _________________________ the third side and ___________________________ . Recall: The area of a trianGle is: � = Median: A line from one vertex to the midpoint of the opposite side. A B C 1) Draw the midpoint of BC, label it M 2) Draw the median from A to BC 3) Calculate the area of rABM and rAMC (MeasurinG any lenGths you need to find the area) Medians of a trianGle ________ its area. Example 2 The area of rXYZ is 48 ��(. A median is drawn from X, and the midpoint of the opposite side is labelled M. What is the area of rXYM? 1. Find the length of line segment MN in each triangle. a) b) c) d) 2. Find the lengths of line segments AD and DE in each triangle. a) b) c) d) 3. The area of rABC is 10 cm2. Calculate the area of each triangle. a) rABD b) rADC 4. Calculate the area of each triangle given the area of rMNQ is 12 cm2. -
Power of a Point and Radical Axis
Power of a Point and Radical Axis Tovi Wen NYC Math Team §1 Power of a Point This handout will cover the topic power of a point, and one of its more powerful uses in radical axes. Definition (Power of a Point) Given a circle ! with center O and radius r, and a point P , the power of P with respect to !, which we will denote as (P; !) is OP 2 − r2. Note that • If P is outside ! then (P; !) is positive. • If P is on ! then (P; !) = 0. • If P is inside ! then (P; !) is negative. This definition does not feel particularly motivated, and therefore when taught at a more elementary level, it is often skipped and replaced by the following nice property. Theorem Let ! be a circle and let P be a point not on !. If a line passing through P meets ! at distinct points A and B then ( PA · PB if P lies outside !; (P; !) = −PA · PB if P lies inside ! Proof. It is not immediately obvious why the quantity PA · PB should be fixed for any line passing through P . Draw another chord of ! passing through P as shown. B A ω P O C M D Recall that opposite angles in a cyclic quadrilateral are supplementary. This gives \P AC = \P DB so as \AP C ≡ \BP D is shared, we have 4P AC ∼ 4P DB. In particular, PA PD = =) PA · PB = PC · P D: PC PB 1 Power of a Point and Radical Axis Tovi Wen We now show this quantity is equal to OP 2 − r2. -
Power of a Point and Ceva's Theorem
Mathematical Problem Solving Power of a Point A rather simple definition of the power of a point with respect to a circle is: Let C be a circle of radius r. The power of a point P with respect to C is given by d2 − r2, where d is the distance of P to the center of the circle. Just to fix ideas, for example, if C is the circle of radius r centered at the origin, and P has coordinates (x, y), then the power of P with respect to this circle is x2 + y2 − r2. If P is a point, C a circle, I’ll write Π(P, C) to denote the power of P with respect to C. Obviously, Π(P, C) > 0 if and only if P is outside the circle, < 0 if and only if P is inside the circle, and 0 if and only if P is on the circle. The significance of this concept is due to the following result. Theorem 1 Let P,X,X0 be collinear points, let C be a circle. If X,X0 lie on C, and either X 6= X0, or if X = X0 6= P and the line containing X and P is tangent to C, then 0 Π(P, C)= PX · PX , where PX,PX0 are “directed lengths;” to be specific: PX · PX0 < 0 if P is strictly between X,X0, > 0 if either X is strictly between P and X0 or X0 is strictly between P and X, 0 if P coincides with X or X0. -
The Euler Line in Non-Euclidean Geometry
California State University, San Bernardino CSUSB ScholarWorks Theses Digitization Project John M. Pfau Library 2003 The Euler Line in non-Euclidean geometry Elena Strzheletska Follow this and additional works at: https://scholarworks.lib.csusb.edu/etd-project Part of the Mathematics Commons Recommended Citation Strzheletska, Elena, "The Euler Line in non-Euclidean geometry" (2003). Theses Digitization Project. 2443. https://scholarworks.lib.csusb.edu/etd-project/2443 This Thesis is brought to you for free and open access by the John M. Pfau Library at CSUSB ScholarWorks. It has been accepted for inclusion in Theses Digitization Project by an authorized administrator of CSUSB ScholarWorks. For more information, please contact [email protected]. THE EULER LINE IN NON-EUCLIDEAN GEOMETRY A Thesis Presented to the Faculty of California State University, San Bernardino In Partial Fulfillment of the Requirements for the Degree Master of Arts in Mathematics by Elena Strzheletska December 2003 THE EULER LINE IN NON-EUCLIDEAN GEOMETRY A Thesis Presented to the Faculty of California State University, San Bernardino by Elena Strzheletska December 2003 Approved by: Robert Stein, Committee Member Susan Addington, Committee Member Peter Williams, Chair Terry Hallett, Department of Mathematics Graduate Coordinator Department of Mathematics ABSTRACT In Euclidean geometry, the circumcenter and the centroid of a nonequilateral triangle determine a line called the Euler line. The orthocenter of the triangle, the point of intersection of the altitudes, also belongs to this line. The main purpose of this thesis is to explore the conditions of the existence and the properties of the Euler line of a triangle in the hyperbolic plane. -
On a Construction of Hagge
Forum Geometricorum b Volume 7 (2007) 231–247. b b FORUM GEOM ISSN 1534-1178 On a Construction of Hagge Christopher J. Bradley and Geoff C. Smith Abstract. In 1907 Hagge constructed a circle associated with each cevian point P of triangle ABC. If P is on the circumcircle this circle degenerates to a straight line through the orthocenter which is parallel to the Wallace-Simson line of P . We give a new proof of Hagge’s result by a method based on reflections. We introduce an axis associated with the construction, and (via an areal anal- ysis) a conic which generalizes the nine-point circle. The precise locus of the orthocenter in a Brocard porism is identified by using Hagge’s theorem as a tool. Other natural loci associated with Hagge’s construction are discussed. 1. Introduction One hundred years ago, Karl Hagge wrote an article in Zeitschrift fur¨ Mathema- tischen und Naturwissenschaftliche Unterricht entitled (in loose translation) “The Fuhrmann and Brocard circles as special cases of a general circle construction” [5]. In this paper he managed to find an elegant extension of the Wallace-Simson theorem when the generating point is not on the circumcircle. Instead of creating a line, one makes a circle through seven important points. In 2 we give a new proof of the correctness of Hagge’s construction, extend and appl§ y the idea in various ways. As a tribute to Hagge’s beautiful insight, we present this work as a cente- nary celebration. Note that the name Hagge is also associated with other circles [6], but here we refer only to the construction just described. -
Prove It Classroom
How to Prove it ClassRoom SHAILESH SHIRALI Euler’s formula for the area of a pedal triangle Given a triangle ABC and a point P in the plane of ABC In this episode of “How (note that P does not have to lie within the triangle), the To Prove It”, we prove pedal triangle of P with respect to ABC is the triangle △ a beautiful and striking whose vertices are the feet of the perpendiculars drawn from formula first found by P to the sides of ABC. See Figure 1. The pedal triangle relates Leonhard Euler; it gives the area of the pedal triangle in a natural way to the parent triangle, and we may wonder of a point with reference whether there is a convenient formula giving the area of the to another triangle. pedal triangle in terms of the parameters of the parent triangle. The great 18th-century mathematician Euler found just such a formula (given in Box 1). It is a compact and pleasing result, and it expresses the area of the pedal triangle in terms of the radius R of the circumcircle of ABC and the △ distance between P and the centre O of the circumcircle. A O: circumcentre of ABC E △ P: arbitrary point Area ( DEF) 1 OP2 F △ = 1 P Area ( ABC) 4 − R2 O △ ( ) B D C Figure 1. Euler's formula for the area of the pedal triangle of an arbitrary point Keywords: Circle theorem, pedal triangle, power of a point, Euler, sine rule, extended sine rule, Wallace-Simson theorem Azim Premji University At Right Angles, July 2018 95 1 A E F P O B D C Figure 2.