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S T P E C N & S O T C S TE G E O M E T R Y GEOMETRY TRIANGLE CENTRES Rajasthan AIR-24 SSC SSC (CGL)-2011 CAT Raja Sir (A K Arya) Income Tax Inspector CDS : 9587067007 (WhatsApp) Chapter 4 Triangle Centres fdlh Hkh triangle ds fy, yxHkx 6100 centres Intensive gSA defined Q. Alice the princess is standing on a side AB of buesa ls 5 Classical centres important gSa ftUgs ge ABC with sides 4, 5 and 6 and jumps on side bl chapter esa detail ls discuss djsaxsaA BC and again jumps on side CA and finally 1. Orthocentre (yEcdsUnz] H) comes back to his original position. The 2. Incentre (vUr% dsUnz] I) smallest distance Alice could have jumped is? jktdqekjh ,fyl] ,d f=Hkqt ftldh Hkqtk,sa vkSj 3. Centroid (dsUnzd] G) ABC 4, 5 lseh- gS fd ,d Hkqtk ij [kM+h gS] ;gk¡ ls og Hkqtk 4. Circumcentre (ifjdsUnz] O) 6 AB BC ij rFkk fQj Hkqtk CA ij lh/kh Nykax yxkrs gq, okfil vius 5. Excentre (ckº; dsUnz] J) izkjafHkd fcUnq ij vk tkrh gSA ,fyl }kjk r; dh xbZ U;wure nwjh Kkr djsaA 1. Orthocentre ¼yEcdsUnz] H½ : Sol. A fdlh triangle ds rhuksa altitudes (ÅapkbZ;ksa) dk Alice intersection point orthocentre dgykrk gSA Stands fdlh vertex ('kh"kZ fcUnw) ls lkeus okyh Hkqtk ij [khapk x;k F E perpendicular (yEc) altitude dgykrk gSA A B D C F E Alice smallest distance cover djrs gq, okfil viuh H original position ij vkrh gS vFkkZr~ og orthic triangle dh perimeter (ifjeki) ds cjkcj distance cover djrh gSA Orthic triangle dh perimeter B D C acosA + bcosB + c cosC BAC + BHC = 1800 a = BC = 4 laiwjd dks.k (Supplementary angles- ) b = AC = 5 BAC = side BC ds opposite vertex dk angle c = AB = 6 BHC = side BC }kjk orthocenter (H) ij cuk;k 2 2 2 b +c -a 3 x;k angle. cosA = = 2bc 4 Proof : a2+c2-b2 9 cosB = = AEHF ,d cyclic quadrilateral ¼ pØh; prqHkqZt ½ gSA 2ac 16 0 0 F = 90 , E = 90 a2+b2-c2 cosC = = 1 FAE + EHF = 1800 2ab 8 FAE = BAC 3 9 105 Perimeter = 4 + 5 + 6 1 = cm. EHF = BHC 4 16 8 16 (vertically opposite angles - f'k"kkZfHkeq[k dks.k) BAC + BHC = 1800 Download our app : NEON CLASSES 1 geometry NEON CLASSES, JAIPUR • 9828728833/34 Orthic triangle 4. Orthic triangle dsoy acute angled triangles esa gh gS rFkk o Triangle ds rhuks altitudes ds feet dks feykus ls cuus possible right angled obtuse angled okyk dgykrk gSA ;g (vf/kd dks.k f=Hkqt) ds fy, orthic triangle ugh cuk;k triangle orthic triangle tk ldrk gSA triangle orthocentre ds fy, pedal triangle Hkh gksrk gSA A Orthic triangle esa cuus okys cyclic quadrilaterals (pdzh; prqHkqZt) AFHE, BFHD, CEHD BCEF, ACDF, ABDE F E Orthic triangle DEF dh inradius and circumradius H Inradius = 2RcosAcosBcosC Circumradius = R/2 Triangle ABC ds vertices ('kh"kZ fcUnw) pedal B C D triangle DEF ds excentres gksrs gSaA vFkkZr~ ABC dh Triangle DEF orthic triangle gSA Hkqtk;sa orthic triangle DEF ds fy, exterior angle cká dks.k lef}Hkktd gksrh gSA 1. Triangle ABC dk orthocentre H, orthic triangle bisectors ( ) DEF dk incentre (vUr% dsUnz) gksrk gSA vFkkZr~ original Vertex A, B rFkk C ls orthic triangle DEF dh Hkqtkvksa triangle ABC ds altitudes orthic triangle DEF ds ij Mkys x, yEc (perpendiculars) ABC ds fy, interior angle bisectors gksrs gSa] vFkkZr~ circumcentre (ifjdsUnz) O ij feyrs gSa vFkkZr~ ABC dh ifjo`r dh f=T;k f=Hkqt dh Hkqtkvksa ∠BEF = ∠BED, ∠ADE = ∠ADF, ∠CFE = ∠CFD circumradius ( ) DEF ij yEcor~ gksrh gSA 2. Triangle ABC esa inscribed fd, tk ldus okys lHkh T A triangles esaa lcls de perimeter (ifjeki) orthic triangle dh gksrh gSA Orthic triangle dh sides = F E (i) R Sin2A = a Cos A O Circumcentre (ii) R Sin2B = b Cos B ¼ifjdsUnz½ (iii) R Sin 2C = c Cos C B D C vr% orthic triangle dh perimeter = Orthocentre and circumcircle ds e/; laca/k & acosA + bcosB + ccosC. A abc cosAcosBcosC Area of orthic triangle DEF = 2R K R = ABC dh circumradius F E H 3. Orthic triangle DEF dk circumcircle original triangle ABC dh Hkqtkvksa ds mid points ls xqtjrk gS rFkk ;g circumcircle, orthocentre ls vertices dh B C nwjh ds e/; fcanqvksa ls Hkh xqtjrk gSA ;g bl triangle dk ij dk gSA nine point circle dgykrk gS] tks fd orthic triangle FK, line AB orthocentre H reflection vr% ,d gS rFkk dk DEF ds fy, circumcircle ¼ifjo`r½ gksrk gSA HBKA kite line segment HK midpoint F gSA Nine point circle dh radius, ABC dh circum radius dh vk/kh gksrh gSA ∠ABH = ∠ABK Like Our FB Page : @neon.classes 2 Web.: www.neonclasses.com rFkk nksuks gS rFkk FE 1 A Orthic DEF JKL similar KJ = 2 A F E K J F E H B D C bl case esa D, E rFkk F altitudes gSaA B D C BFC ~ BDA L (B common gS rFkk ∠F = ∠D = 900, vr% AA property ls) AEB ~ AFC ⇨ CDA ~ CEB Concurrent Lines BF BC AC AB CD CA vr% = ; = ; = 3 ;k 3 ls vf/kd lines tks fdlh ,d point ij intersect BD BA AF AC CE BC djrh gS vFkkZr~ fdlh ,d gh fcanw ij vkdj feyrh gS] bu rhuksa equations dks multiply djus ij] concurrent lines dgykrh gSA rFkk og fcanw ftl ij ;s BF . AE . CD = BC . AB . CA = 1 vkil esa dkVrh gSa BD AF CE BA AC BC lines point of concurrency dks djus ij dgykrk gSA Left Hand Side rearrange AF BD CE Line a . = 1 Eg.: FB DC EA vr% ABC esa rhuksa altitudes ,d single point ij feysaxsa Line b m ftls orthocentre (yEcdsUnz) dgrs gSaA Incentre ds fy,& Line c fdlh ABC esa incentre exist djrk gS bldk proof dk gS ;gka fcanw m, point of concurrency dgykrk gSA Ceva's theorem trigonometric version ftlds vuqlkj AD, BE rFkk CF rHkh concur djsaxh tc& fdlh triangle ds fy, fofHkUu lines (vyx&vyx ds vuqlkj fdlh ,d fcanw ij vkil esa dkVrh gSA Sin ∠ BAD . Sin ∠CBE . Sin ∠A CF = 1 concepts ) Sin ∠ABE Sin ∠BCF Sin ∠CAD bl fcanw dks mu lines ds vuqlkj vyx&vyx centres A (Eg. orthocentre, incentre, centroid, circumcentre, θ θ excentre etc.) ds :Ik esa define fd;k x;k gSA 2 2 F E Ceva'a theorem fdUgh Hkh rhu lines ds fy, concurrence crkrh gS vr% ;g orthocentre, incentre rFkk circumcentre ds fy, Hkh true gS rFkk Ceva's theorem dh lgk;rk ls ge bu centres dk B D C existence prove dj ldrs gSaA Incentre ds case esa D, E rFkk F angle bisectors ds feet gSa vr% Proof of existence ∠BAD = ∠CAD, ∠ABE = ∠CBE rFkk ∠ACF = ∠BCF fdlh Hkh esa yEcdsUnz triangle orthocentre ( ) exist vr% Sin ∠ BAD . Sin ∠ABE . Sin ∠A CF = 1 : 1.1 = 1 djrk gS bldk lcls simple proof Ceva's theorem gSA Sin ∠CAD Sin ∠CBE Sin ∠BCF Left Hand Side dks rearrange djus ij] Ceva's theorem ds vuqlkj] Sin ∠BAD Sin ∠CBE Sin ∠ACF . = 1 Rkhu lines AD, BE rFkk CF rHkh concur gksaxh vFkkZr~ fdlh Sin ∠ABE Sin ∠BCF Sin ∠CAD ,d fcanw ij feysaxh tc & vr% rhuksa angle bisectors ,d point ij feysaxsa ftls AF. BD . CE = 1, vUr%dsUnz dgrs gSaA FB DC EA incentre ( ) Download our app : NEON CLASSES 3 geometry NEON CLASSES, JAIPUR • 9828728833/34 Orthocentric System ADB esa Triangle ABC rFkk bldk orthocentre H feydj ,d AB > AD ......... (i) orthocentric system cukrs gSa vFkkZr~ fdUgh Hkh rhu BEC esa points dks feykus ls cuus okys triangle ds fy, pkSFkk BC > BE .........(ii) gksrk gSA point orthocentre CFA esa A AC > CF ..........(iii) adding (i), (ii), (iii) we get F E AB+BC+AC > AD+BE+CF H F fdlh triangle ABC esa altitudes dk vuqikr %& 1 1 1 Area of ABC = 2 ABCF = 2 BC AD = 2 AC BE 1 1 1 AB : BC : AC = : : CF AD BE B D C Equilateral triangle ¼leckgq f=Hkqt½ esa rhuksa Hkqtk,sa leku gksrh gS] vr% altitudes ¼yEc½ Hkh leku gksrs gSaA F ABC dk orthocentre = H ABH dk orthocentre = C Position of Orthocentre- BCH dk orthocentre = A (i) Acute angled triangle (U;wu dks.k f=Hkqt) :- ACH dk orthocentre = B blesa orthocentre f=Hkqt ds vUnj fLFkr gksrk gSA F AH.HD = BH.HE = CH.HF A Orthocentre, altitudes dks ftu nks lengths esa divide djrk gS mudk product ges'kk ,d constant gksrk gSA H Similarly, (i) AD.DH = BD.DC (ii) BE.EH = AE.EC B C 0 (iii) CF.FH = AF.FB A, B, C rhuksa 90 ls de gSA AF BD CE (ii) Right angled triangle (ledks.k f=Hkqt) F = 1 FB DC EA F Pair of similar triangles A DBH ~ EHA DHC ~ FHA FHB ~ EHC F Sum of three altitudes of a triangle is less than sum of three sides of a triangle. BH C fdlh f=Hkqt ds rhuksa altitudes ¼yEcksa½ dk ;ksx rhuksa Hkqtkvksa Orthocentre ds ;ksx ls de gksrk gSA fdlh ledks.k f=Hkqt esa] Right angled triangle esaa orthocentre ledks.k okys ij fLFkr gksrk gSA Hypotenuse ¼d.kZ½ > altitude ¼yEc½ vertex Like Our FB Page : @neon.classes 4 Web.: www.neonclasses.com (iii) Obtuse angled triangle (vf/kd dks.k f=Hkqt) (2) Incentre ¼vUr% dsUnz] I ½ Obtuse angled triangle esa orthocenter vf/kd dks.k okys 'kh"kZ dh rjQ f=Hkqt ls ckgj fudy tkrk gSA f=Hkqt ds rhuksa angles ds angle bisectors (dks.k 0 lef}Hkktd) dk intersection point (dVku fcUnw) B > 90 , B obtuse angle gSA Incentre dgykrk gSA A A 2 / 2 / E I r F C B r D /2 r /2 H /2 /2 B C 0 ABC + AHC = 180 Incentre ls rhuksa sides dh nwjh leku gksrh gS bl nwjh dks A vUr% o`r dh f=T;k dgrs gSa rFkk blls cuus inradius (r ) E okyk o`r incircle (vUr%o`r) dgykrk gSA F Incentre ges'kk f=Hkqt ds vUnj gh fLFkr gksrk gSA rFkk F C B okLro esa ml f=Hkqt ds medial triangle ds vUnj fLFkr gksrk gSA A D H K I BAC + KHC = 1800 Proof /2 /2 EBA ~ DBH B C 0 E = D = 90 fdlh Hkqtk }kjk I ij cuus okyk angle rFkk Hkqtk ds EBA = DBH (vertically opposite angles) opposite vertex ij cuus okys angle dk laca/k& EAB =DHB (similar triangles) 0 = 90 + KHC + DHB = 180 (Straight line ij cus angles) 2 EAB = BAC Proof : 0 KHC + BAC = 180 BIC esa ⇒ + + = 1800 .........
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    High School Geometry This course covers the topics shown below. Students navigate learning paths based on their level of readiness. Curriculum Arithmetic and Algebra Review (151 topics) Fractions and Decimals (28 topics) Factors Greatest common factor of 2 numbers Equivalent fractions Simplifying a fraction Division involving zero Introduction to addition or subtraction of fractions with different denominators Addition or subtraction of fractions with different denominators Product of a unit fraction and a whole number Product of a fraction and a whole number: Problem type 1 Fraction multiplication Product of a fraction and a whole number: Problem type 2 The reciprocal of a number Division involving a whole number and a fraction Fraction division Complex fraction without variables: Problem type 1 Decimal place value: Tenths and hundredths Rounding decimals Introduction to ordering decimals Using a calculator to convert a fraction to a rounded decimal Addition of aligned decimals Decimal subtraction: Basic Decimal subtraction: Advanced Word problem with addition or subtraction of 2 decimals Multiplication of a decimal by a power of ten Introduction to decimal multiplication Multiplying a decimal by a whole number Word problem with multiple decimal operations: Problem type 1 Converting a fraction to a terminating decimal: Basic Signed Numbers (14 topics) Plotting integers on a number line Ordering integers Absolute value of a number Integer addition: Problem type 1 Integer addition: Problem type 2 Integer subtraction: Problem type 1 Integer
  • Orthocenters of Triangles in the N-Dimensional Space

    Orthocenters of Triangles in the N-Dimensional Space

    Divulgaciones Matemáticas Vol. 17 No. 2 (2016), pp. 114 Orthocenters of triangles in the n-dimensional space Ortocentros para triángulos en el espacio n-dimensional Horst Martini([email protected]) Fakultät für Mathematik, TU Chemnitz, 09107 Chemnitz, Germany Wilson Pacheco ([email protected]) Aljadis Varela ([email protected]) John Vargas ([email protected]) Departamento de Matematicas Facultad Experimental de Ciencias Universidad del Zulia Maracaibo - Venezuela Abstract We present a way to dene a set of orthocenters for a triangle in the n-dimensional space n R , and we show some analogies between these orthocenters and the classical orthocenter of a triangle in the Euclidean plane. We also dene a substitute of the orthocenter for tetra- hedra which we call G−orthocenter. We show that the G−orthocenter of a tetrahedron has some properties similar to those of the classical orthocenter of a triangle. Key words and phrases: orthocenter, triangle, tetrahedron, orthocentric system, Feuerbach sphere. Resumen Presentamos una manera de denir un conjunto de ortocentros de un triángulo en el n espacio n-dimensional R , y mostramos algunas analogías entre estos ortocentros y el or- tocentro clásico de un triángulo en el plano euclidiano. También denimos un sustituto del ortocentro para tetraedros que llamamos G−ortocentro. Se demuestra que el G−ortocentro de un tetraedro tiene algunas propiedades similares a los del ortocentro clásico de un trián- gulo. Palabras y frases clave: ortocentro, triángulo, tetraedro, sistema ortocéntrico, esfera de Feuerbach. 1 Introduction In the Euclidean plane, the orthocenter H of a triangle 4ABC is known as the unique point where the altitudes of the triangle intersect, i.e., the point at which the three lines perpendicular to Received 20/07/16.