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A Synthetic Proof of Goormaghtigh’s Generalization of

Y, Z dividing OA, OB, OC respectively in the ratio OX OA = OY OB = OZ OC = t (†) The lines AX, BY, CZ are concurrent at the isogonal conjugate of the point P on the Euler line dividing OH in the ratio OP: PH=1:2t Proof Let the isogonal line of AX (with respect to angle A) intersect OA at X The triangles OAX and OX A are similar It



AUXILIARY DEVICES - Siemens

OA One normally open and one normally closed OB one normally open OC one normally closed OS one normally closed late opening OR one normally open early closing OG two normally open OH two normally closed OJ same as OA, but NC contact is late opening OV same as OA, but NO contact is early closing 2 pilot light units for BA9s bulbs



A Note on the Hervey Point of a Complete Quadrilateral

OH= OA+OB+OC It is easy to see that the resultant AHof the two vectors OBand OCis perpen-dicular to the side BC Therefore the line AH is the altitude drawn through the vertex A and, more generally, the point H is the point of concurrence of the three altitudes The usual property of the orthocenter is recovered



Inscribing a similar triangle using spiral similarity

OH= OA 0+ OB0 + OC Now we know that the sum of the vectors from the symmedian point to the vertices of its pedal triangle is zero, and the sum of the vectors from the circumcenter to the vertices of its pedal triangle is the vector from it to the orthocenter, i e , in the direction of the Euler line, which contains both points



(OA OB OC) ( )( ) ( )( ) AH⋅BC= OH−OA⋅OB−OC= OB+OCOB−OC=OB

(1)og (oa ob oc) oh 3 1 3 1 = + + = を用いて証明 (2) oa=ob=ocに注意すると ( )( ) ( )( ) 0 2 2 ah⋅bc= oh−oa⋅ob−oc= ob+ocob−oc=ob −oc = 同様にbh⋅ca=ch⋅ab=0より、各々のパェダャが直交するので点hは垂心 ＜寸評＞ 外心oン垂心hン重心gが一直線上にあり、線分比についてoh:hg=1:2



Exercise 13(A) Page: 158 - BYJUS

OA 2= AH 2+ OH = AH + AE OC 2= CG + OG2 = EB2 + HD OB 2= EO + BE 2= AH + BE2 OD 2= HD 2+ OH = HD + AE2 Adding these equalities we get: OA 2+ OC 2= AH2 + HD + AE + EB OB 2+ OD = AH2 + HD2 + AE + EB From which we prove that for any point within the rectangle there is the relation OA 2+ OC = OB2 + OD2 Hence Proved 9



1 Yellow Dragon 08 - Queens College, Hong Kong

Title: Microsoft Word - 1_Yellow Dragon 08 doc Author: YUE Created Date: 9/3/2015 5:22:09 PM



BC AC AB 1

oh og 0 hg o C,1 ; B,1 ; A,1 G ABC G





تمارين و حلول المعلم في المستوى

(3) oA + →oB + oC = 3 → oG → وأ : ج ˚ ا (4) → →oH + →oB + oC = oH : ˚˜ (2) لاB#ا C#D oH = 3 → oG → نأ - ˚ #˘ (4) و (3) C#D نذ إ " #ﻡ H و G و o E"˚ا و ا سر ا سر

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?? ??M'?????? ?? B ??? A??? ???? ???? ???? M -???? ???? ????? ABC ????

ADCB 2-??????? ????? ??????

??→AE = ??→AD + ??→AC ???? E 3- ?????? ????? . ????? ????C ??? B ??????? ???? ???? ABC 4-?????? ???? ??? ???? :G ???? ? o ????? ???? ?????? ABCD ???? ??→AG = 1 2 ??→AB + 3 2 ??→AD ??→DG = ??→Ao ?? ??? ? G 1-????? ??→AH = 1 2 ??→AB - 5 2 ??→AD ???? H 2-????? ??→HG= 4 ??→AD ?? ??? ? H ????? ???? ?????? ABCD ???? ??→CM= ??→CA + ??→CD : ???? ???? M ? [BM] ????? ?? A 1-?? ??? [BE] ????? ?? D ?? ??? [AB] ????? I ? ???? ABC ???? ??→BF = 2 ??→IJ ???? F 2-????? F ?? ??→BC??? ??????? ??? ???? C 3- ???? ?? ??? ??→oB + ??→oC = ??→oD 1-?? ??? ??→oH = ??→oA + ??→oB + ??→oC ???? H 2- ????? (AH) ? (BC) ?? ?????? ? ??→AH = ??→oD 3-?? ???

ABC ?????? ??????? H ???? ????

ABC ?????? ??? ???? G 4-????

??→oA + ??→oB + ??→oC = 3 ??→oG ?? ??? ??→AD = ??→BC ? ??: ????? ?????? ABCD ??????? ? ??→AE = ??→AD + ??→AC 3- ????? ??→AD + ??→DE = ??→AD × ??→AC ? ?? ??→DE = ??→EC ??? ??? ? ????? ?????? ADEC ??????? E ? ???? ??????? ????? ?C ???B ??? ???? ???? ????ABC 4-?????? ???? ??→BC ??? ??????? ??? ????ABC ?????? ???? ?? ??→BC ??? ????D ?? A ???? (1) ?????? ??? ?? ????? ??→BC ??? ???? C ????? ?? B ?? C ?? B ???? ????? ??? ?? ????? ??????ADEC (3) ?????? ??? ?? ???? ????? ??→AE = ??→AD + ??→AC ????? ?? ??→AC + ??→CE = ??→AD + ??→AC ? ??: ??→AD = ??→CE ? ?? ??→AD = ??→BC ????? ? ??→CE = ??→BC ? ?? ??→BC ??? ???? E ?? C ???? ??????? ? ??→AG = 1 2 ??→AB + 3 2 ??→AD 1- ????? ??→AD + ??→DG = 1 2 ??→AB + 3 2 ??→AD ??→DG = 1 2 ??→AB + 1 2 ??→AD 1 2 ??→AB + ??→AD) ? ???? ?????? ? ?:ABCD ?? ??? ??→AB + ??→AD = ??→AC ??→DG = 1 2 ??→AC 1 2 (2 ??→Ao) ??→Ao ??→DG = C E B A D o G C D B A ??→AH = 1 2 ??→AB 2- ????? ??→AG + ??→GH = 1 2 ??→AB - 5 2 ??→AD 1 2 ??→AB + 3 2 ??→AD + ??→GH = 1 2 ??→AB - 5 2 ??→AD ??→GH = - 5 2 ??→AD - 3 2 ??→AD 8 2 ??→AD = - 4 ??→AD ??→HG = 4 ??→AD ??→Do = ??→Ao ????? (1) 3-?????? ??? ????? ?????? DGAo ??? ??→oG = ??→AD ? ?? ??→HG= 4 ??→AD (2) ?????? ??? ? ??→HG = 4 ??→AD = 4 ??→oG ? ?? ??→CM = ??→CA + ??→CD 1- ????? ??→CA + ??→AM = ??→CA + ??→BA ??→AM = ??→BA ??????? ? ??→MA = ??→AB ?? [MB] ????? A ? ?? 2- ??→BD??? ??????? ??? ???? D ???? ?? E ?? ??? ??→DE = ??→BD ??→DB + ??→BE = ??→BD ??→BE = 2 ??→BD ??→BD = 1 2 ??→BE [BE] ????? ?? D ??????? ? o G C D B A H E M D C B A ??→BC ??? ??????? ??? ????I? ??? J 1- ??→BC = ??→IJ ? ??: ????? ?????? (IJCB) ??????? J ???? ??????? ?????? ??→BF = 2 ??→IJ ???? F 2- ????? F ?? ??→BC = ??→CF ?? ???? ?? ???: ??→BF = 2 ??→IJ ????? (2) ?????? ??? ?? ??→BC + ??→CF = 2 ??→BC ?? ??→CF = 2 ??→BC - ??→BC ??→CF = ??→BC (1) ??→IB = ??→JC ? ????? ?????? IJCB ??? ??→BC = ??→IJ 4-?? ??? ??→AI =quotesdbs_dbs15.pdfusesText_21