Isosceles Triangle Examples in Real Life
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Angle Chasing
Angle Chasing Ray Li June 12, 2017 1 Facts you should know 1. Let ABC be a triangle and extend BC past C to D: Show that \ACD = \BAC + \ABC: 2. Let ABC be a triangle with \C = 90: Show that the circumcenter is the midpoint of AB: 3. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove that H is the incenter of 4DEF . 4. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove (i) that A; E; F; H lie on a circle diameter AH and (ii) that B; E; F; C lie on a circle with diameter BC. 5. Let ABC be a triangle with circumcenter O and orthocenter H: Show that \BAH = \CAO: 6. Let ABC be a triangle with circumcenter O and orthocenter H and let AH and AO meet the circumcircle at D and E, respectively. Show (i) that H and D are symmetric with respect to BC; and (ii) that H and E are symmetric with respect to the midpoint BC: 7. Let ABC be a triangle with altitudes AD; BE; and CF: Let M be the midpoint of side BC. Show that ME and MF are tangent to the circumcircle of AEF: 8. Let ABC be a triangle with incenter I, A-excenter Ia, and D the midpoint of arc BC not containing A on the circumcircle. Show that DI = DIa = DB = DC: 9. Let ABC be a triangle with incenter I and D the midpoint of arc BC not containing A on the circumcircle. -
The Construction, by Euclid, of the Regular Pentagon
THE CONSTRUCTION, BY EUCLID, OF THE REGULAR PENTAGON Jo˜ao Bosco Pitombeira de CARVALHO Instituto de Matem´atica, Universidade Federal do Rio de Janeiro, Cidade Universit´aria, Ilha do Fund˜ao, Rio de Janeiro, Brazil. e-mail: [email protected] ABSTRACT We present a modern account of Ptolemy’s construction of the regular pentagon, as found in a well-known book on the history of ancient mathematics (Aaboe [1]), and discuss how anachronistic it is from a historical point of view. We then carefully present Euclid’s original construction of the regular pentagon, which shows the power of the method of equivalence of areas. We also propose how to use the ideas of this paper in several contexts. Key-words: Regular pentagon, regular constructible polygons, history of Greek mathe- matics, equivalence of areas in Greek mathematics. 1 Introduction This paper presents Euclid’s construction of the regular pentagon, a highlight of the Elements, comparing it with the widely known construction of Ptolemy, as presented by Aaboe [1]. This gives rise to a discussion on how to view Greek mathematics and shows the care on must have when adopting adapting ancient mathematics to modern styles of presentation, in order to preserve not only content but the very way ancient mathematicians thought and viewed mathematics. 1 The material here presented can be used for several purposes. First of all, in courses for prospective teachers interested in using historical sources in their classrooms. In several places, for example Brazil, the history of mathematics is becoming commonplace in the curricula of courses for prospective teachers, and so one needs materials that will awaken awareness of the need to approach ancient mathematics as much as possible in its own terms, and not in some pasteurized downgraded versions. -
Refer to the Figure. 1. If Name Two Congruent Angles. SOLUTION: Isosceles Triangle Theorem States That If Two Sides of T
4-6 Isosceles and Equilateral Triangles Refer to the figure. 1. If name two congruent angles. SOLUTION: Isosceles Triangle Theorem states that if two sides of the triangle are congruent, then the angles opposite those sides are congruent. Therefore, in triangle ABC, ANSWER: BAC and BCA 2. If EAC ECA, name two congruent segments. SOLUTION: Converse of Isosceles Triangle Theorem states that if two angles of a triangle are congruent, then the sides opposite those angles are congruent. Therefore, in triangle EAC, ANSWER: Find each measure. 3. FH SOLUTION: By the Triangle Angle-Sum Theorem, Since the measures of all the three angles are 60°; the triangle must be equiangular. All the equiangular triangles are equilateral. Therefore, FH = GH = 12. ANSWER: 12 eSolutions4. m ManualMRP - Powered by Cognero Page 1 SOLUTION: Since all the sides are congruent, is an equilateral triangle. Each angle of an equilateral triangle measures 60°. Therefore, m MRP = 60°. ANSWER: 60 SENSE-MAKING Find the value of each variable. 5. SOLUTION: In the figure, . Therefore, triangle RST is an isosceles triangle. By the Converse of Isosceles Triangle Theorem, That is, . ANSWER: 12 6. SOLUTION: In the figure, Therefore, triangle WXY is an isosceles triangle. By the Isosceles Triangle Theorem, . ANSWER: 16 7. PROOF Write a two-column proof. Given: is isosceles; bisects ABC. Prove: SOLUTION: ANSWER: 8. ROLLER COASTERS The roller coaster track appears to be composed of congruent triangles. A portion of the track is shown. a. If and are perpendicular to is isosceles with base , and prove that b. If VR = 2.5 meters and QR = 2 meters, find the distance between and Explain your reasoning. -
Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 1
Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 1 Similar Quadrilaterals Authors Guangqi Cui, Akshaj Kadaveru, Joshua Lee, Sagar Maheshwari Special thanks to Cosmin Pohoata and the AMSP Cornell 2014 Geometric Proofs Class B0 C0 B A A0 D0 C D Additional thanks to Justin Stevens and David Altizio for the LATEX Template Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 2 Contents 1 Introduction 3 2 Interesting Property 4 3 Example Problems 5 4 Practice Problems 11 Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 3 1 Introduction Similar quadrilaterals are a very useful but relatively unknown tool used to solve olympiad geometry problems. It usually goes unnoticed due to the confinement of geometric education to the geometry of the triangle and other conventional methods of problem solving. Also, it is only in very special cases where pairs of similar quadrilaterals exist, and proofs using these qualities usually shorten what would have otherwise been an unnecessarily long proof. The most common method of finding such quadrilaterals involves finding one pair of adjacent sides with identical ratios, and three pairs of congruent angles. We will call this SSAAA Similarity. 0 0 0 0 Example 1.1. (SSAAA Similarity) Two quadrilaterals ABCD and A B C D satisfy \A = AB BC A0, B = B0, C = C0, and = . Show that ABCD and A0B0C0D0 are similar. \ \ \ \ \ A0B0 B0C0 B0 C0 B A A0 D0 C D 0 0 0 0 0 0 Solution. Notice 4ABC and 4A B C are similar from SAS similarity. Therefore \C A D = 0 0 0 0 0 0 0 0 0 0 \A − \B A C = \A − \BAC = \CAD. -
An Example on Five Classical Centres of a Right Angled Triangle
An example on five classical centres of a right angled triangle Yue Kwok Choy Given O(0, 0), A(12, 0), B(0, 5) . The aim of this small article is to find the co-ordinates of the five classical centres of the ∆ OAB and other related points of interest. We choose a right-angled triangle for simplicity. (1) Orthocenter: The three altitudes of a triangle meet in one point called the orthocenter. (Altitudes are perpendicular lines from vertices to the opposite sides of the triangles.) If the triangle is obtuse, the orthocenter is outside the triangle. If it is a right triangle, the orthocenter is the vertex which is the right angle. Conclusion : Simple, the orthocenter = O(0, 0) . (2) Circum-center: The three perpendicular bisectors of the sides of a triangle meet in one point called the circumcenter. It is the center of the circumcircle, the circle circumscribed about the triangle. If the triangle is obtuse, then the circumcenter is outside the triangle. If it is a right triangle, then the circumcenter is the midpoint of the hypotenuse. (By the theorem of angle in semi-circle as in the diagram.) Conclusion : the circum -centre, E = ͥͦͮͤ ͤͮͩ ʠ ͦ , ͦ ʡ Ɣ ʚ6, 2.5ʛ 1 Exercise 1: (a) Check that the circum -circle above is given by: ͦ ͦ ͦ. ʚx Ǝ 6ʛ ƍ ʚy Ǝ 2.5ʛ Ɣ 6.5 (b) Show that t he area of the triangle with sides a, b, c and angles A, B, C is ΏΐΑ ͦ , where R is the radius of the circum -circle . -
Chapter 1. the Medial Triangle 2
Chapter 1. The Medial Triangle 2 The triangle formed by joining the midpoints of the sides of a given triangle is called the me- dial triangle. Let A1B1C1 be the medial trian- gle of the triangle ABC in Figure 1. The sides of A1B1C1 are parallel to the sides of ABC and 1 half the lengths. So A B C is the area of 1 1 1 4 ABC. Figure 1: In fact area(AC1B1) = area(A1B1C1) = area(C1BA1) 1 = area(B A C) = area(ABC): 1 1 4 Figure 2: The quadrilaterals AC1A1B1 and C1BA1B1 are parallelograms. Thus the line segments AA1 and C1B1 bisect one another, and the line segments BB1 and CA1 bisect one another. (Figure 2) Figure 3: Thus the medians of A1B1C1 lie along the medians of ABC, so both tri- angles A1B1C1 and ABC have the same centroid G. 3 Now draw the altitudes of A1B1C1 from vertices A1 and C1. (Figure 3) These altitudes are perpendicular bisectors of the sides BC and AB of the triangle ABC so they intersect at O, the circumcentre of ABC. Thus the orthocentre of A1B1C1 coincides with the circumcentre of ABC. Let H be the orthocentre of the triangle ABC, that is the point of intersection of the altitudes of ABC. Two of these altitudes AA2 and BB2 are shown. (Figure 4) Since O is the orthocen- tre of A1B1C1 and H is the orthocentre of ABC then jAHj = 2jA1Oj Figure 4: . The centroid G of ABC lies on AA1 and jAGj = 2jGA1j . We also have AA2kOA1, since O is the orthocentre of A1B1C1. -
Day 7 Classwork for Constructing a Regular Hexagon
Classwork for Constructing a Regular Hexagon, Inscribed Shapes, and Isosceles Triangle I. Review 1) Construct a perpendicular bisector to 퐴퐵̅̅̅̅. 2) Construct a perpendicular line through point P. P • A B 3) Construct an equilateral triangle with length CD. (LT 1c) C D II. Construct a Regular Hexagon (LT 1c) Hexagon A six-sided polygon Regular Hexagon A six-sided polygon with all sides the same length and all interior angles the same measure Begin by constructing an equilateral with the segment below. We can construct a regular hexagon by observing that it is made up of Then construct 5 more equilateral triangles with anchor point on point R 6 adjacent (sharing a side) equilateral for one arc and anchor point on the most recent point of intersection for the second ( intersecting) arc. triangles. R III. Construct a Regular Hexagon Inscribed in a Circle Inscribed (in a Circle) All vertices of the inscribed shape are points on the circle Inscribed Hexagon Hexagon with all 6 vertex points on a circle Construct a regular hexagon with side length 퐴퐵̅̅̅̅: A B 1) Copy the length of 퐴퐵̅̅̅̅ onto the compass. 2) Place metal tip of compass on point C and construct a circle. 3) Keep the same length on the compass. 4) Mark any point (randomly) on the circle. C 5) Place the metal tip on the randomly marked • point and mark an arc on the circle. 6) Lift the compass and place the metal tip on the last arc mark, and mark new arc on the circle. 7) Repeat until the arc mark lands on the original point. -
Lemmas in Euclidean Geometry1 Yufei Zhao [email protected]
IMO Training 2007 Lemmas in Euclidean Geometry Yufei Zhao Lemmas in Euclidean Geometry1 Yufei Zhao [email protected] 1. Construction of the symmedian. Let ABC be a triangle and Γ its circumcircle. Let the tangent to Γ at B and C meet at D. Then AD coincides with a symmedian of △ABC. (The symmedian is the reflection of the median across the angle bisector, all through the same vertex.) A A A O B C M B B F C E M' C P D D Q D We give three proofs. The first proof is a straightforward computation using Sine Law. The second proof uses similar triangles. The third proof uses projective geometry. First proof. Let the reflection of AD across the angle bisector of ∠BAC meet BC at M ′. Then ′ ′ sin ∠BAM ′ ′ BM AM sin ∠ABC sin ∠BAM sin ∠ABD sin ∠CAD sin ∠ABD CD AD ′ = ′ sin ∠CAM ′ = ∠ ∠ ′ = ∠ ∠ = = 1 M C AM sin ∠ACB sin ACD sin CAM sin ACD sin BAD AD BD Therefore, AM ′ is the median, and thus AD is the symmedian. Second proof. Let O be the circumcenter of ABC and let ω be the circle centered at D with radius DB. Let lines AB and AC meet ω at P and Q, respectively. Since ∠PBQ = ∠BQC + ∠BAC = 1 ∠ ∠ ◦ 2 ( BDC + DOC) = 90 , we see that PQ is a diameter of ω and hence passes through D. Since ∠ABC = ∠AQP and ∠ACB = ∠AP Q, we see that triangles ABC and AQP are similar. If M is the midpoint of BC, noting that D is the midpoint of QP , the similarity implies that ∠BAM = ∠QAD, from which the result follows. -
Isosceles Triangles on a Geoboard
Isosceles Triangles on a Geoboard About Illustrations: Illustrations of the Standards for Mathematical Practice (SMP) consist of several pieces, including a mathematics task, student dialogue, mathematical overview, teacher reflection questions, and student materials. While the primary use of Illustrations is for teacher learning about the SMP, some components may be used in the classroom with students. These include the mathematics task, student dialogue, and student materials. For additional Illustrations or to learn about a professional development curriculum centered around the use of Illustrations, please visit mathpractices.edc.org. About the Isosceles Triangles on a Geoboard Illustration: This Illustration’s student dialogue shows the conversation among three students who are looking for how many isosceles triangles can be constructed on a geoboard given only one side. After exploring several cases where the known side is one of the congruent legs, students then use the known side as the base and make an interesting observation. Highlighted Standard(s) for Mathematical Practice (MP) MP 1: Make sense of problems and persevere in solving them. MP 5: Use appropriate tools strategically. MP 7: Look for and make use of structure. Target Grade Level: Grades 6–7 Target Content Domain: Geometry Highlighted Standard(s) for Mathematical Content 7.G.A.2 Draw (freehand, with ruler and protractor, and with technology) geometric shapes with given conditions. Focus on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no triangle. Math Topic Keywords: isosceles triangles, geoboard © 2016 by Education Development Center. Isosceles Triangles on a Geoboard is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. -
1. EUCLIDEAN GEOMETRY 1.3. Triangle Centres
1. EUCLIDEAN GEOMETRY 1.3. Triangle Centres Triangle Centres The humble triangle, with its three vertices and three sides, is actually a very remarkable object. For evidence of this fact, just look at the Encyclopedia of Triangle Centers1 on the internet, which catalogues literally thousands of special points that every triangle has. We’ll only be looking at the big four — namely, the circumcentre, the incentre, the orthocentre, and the centroid. While exploring these constructions, we’ll need all of our newfound geometric knowledge from the previous lecture, so let’s have a quick recap. Triangles – Congruence : There are four simple rules to determine whether or not two triangles are congruent. They are SSS, SAS, ASA and RHS. – Similarity : There are also three simple rules to determine whether or not two triangles are similar. They are AAA, PPP and PAP. – Midpoint Theorem : Let ABC be a triangle where the midpoints of the sides BC, CA, AB are X, Y, Z, respectively. Then the four triangles AZY, ZBX, YXC and XYZ are all congruent to each other and similar to triangle ABC. Circles – The diameter of a circle subtends an angle of 90◦. In other words, if AB is the diameter of a circle and C is a point on the circle, then \ACB = 90◦. – The angle subtended by a chord at the centre is twice the angle subtended at the circumference, on the same side. In other words, if AB is a chord of a circle with centre O and C is a point on the circle on the same side of AB as O, then \AOB = 2\ACB. -
Singularities of Centre Symmetry Sets 1 Introduction
Singularities of centre symmetry sets P.J.Giblin, V.M.Zakalyukin 1 1 Introduction We study the singularities of envelopes of families of chords (special straight lines) intrin- sically related to a hypersurface M ⊂ Rn embedded into an affine space. The origin of this investigation is the paper [10] of Janeczko. He described a general- ization of central symmetry in which a single point—the centre of symmetry—is replaced by the bifurcation set of a certain family of ratios. In [6] Giblin and Holtom gave an alternative description, as follows. For a hypersurface M ⊂ Rn we consider pairs of points at which the tangent hy- perplanes are parallel, and in particular take the family of chords (regarded as infinite lines) joining these pairs. For n = 2 these chords will always have an envelope, and this envelope is called the centre symmetry set (CSS) of the curve M. When M is convex (the case considered in [10]) the envelope is quite simple to describe, and has cusps as its only generic singularities. In [6] the case when M is not convex is considered, and there the envelope acquires extra components, and singularities resembling boundary singularities of Arnold. This is because, arbitrarily close to an ordinary inflexion, there are pairs of points on the curve with parallel tangents. The corresponding chords have an envelope with a limit point at the inflexion itself. When n = 3 we obtain a 2-parameter family of lines in R3, which may or may not have a (real) envelope. The real part of the envelope is again called the CSS of M. -
Centroid and Moment of Inertia 4.1 Centre of Gravity 4.2 Centroid
MODULE 4 Centroid and Moment of Inertia 4.1 Centre of Gravity Everybody is attracted towards the centre of the earth due gravity. The force of attraction is proportional to mass of the body. Everybody consists of innumerable particles, however the entire weight of a body is assumed to act through a single point and such a single point is called centre of gravity. Every body has one and only centre of gravity. 4.2 Centroid In case of plane areas (bodies with negligible thickness) such as a triangle quadrilateral, circle etc., the total area is assumed to be concentrated at a single point and such a single point is called centroid of the plane area. The term centre of gravity and centroid has the same meaning but the following d ifferences. 1. Centre of gravity refer to bodies with mass and weight whereas, centroid refers to plane areas. 2. centre of gravity is a point is a point in a body through which the weight acts vertically downwards irrespective of the position, whereas the centroid is a point in a plane area such that the moment of areas about an axis through the centroid is zero Plane area ‘A’ G g2 g1 d2 d1 a2 a1 Note: In the discussion on centroid, the area of any plane figure is assumed as a force equivalent to the centroid referring to the above figure G is said to be the centroid of the plane area A as long as a1d1 – a2 d2 = 0. 4.3 Location of centroid of plane areas Y Plane area X G Y The position of centroid of a plane area should be specified or calculated with respect to some reference axis i.e.