tag:blogger.com,1999:blog-6933544261975483399.post4862764278177311502..comments2024-09-10T22:02:29.582-07:00Comments on GoGeometry.com (Problem Solutions): Bottema's Theorem: Triangle and SquaresAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-57023875366750314792017-08-25T21:21:23.395-07:002017-08-25T21:21:23.395-07:00https://goo.gl/photos/5ZVEeXc42HDNTC6b7
Let P, N,...https://goo.gl/photos/5ZVEeXc42HDNTC6b7<br /><br />Let P, N, K and Q are the projection of points E, M, B and F over AC ( see sketch)<br />Let DC meet AG at R<br />1,2 Since M is the midpoint of EF so N is the midpoint of PQ<br />Observe that triangle BCK and CFQ are congruent ..( case ASA) so CQ=BK and FQ=CK<br />Simillarly BAK are congruent to AEP ( case ASA) so AP=BK=CQ and EP=AN<br />N is the midpoint of PQ and AP=BK=CQ => N is the midpoint of AC<br />In trapezoid EFQP we have MN= ½(EP+FQ)= ½(AK+KC)= 1/2AC<br />So triangle AMC is an isoceles right triangle and position of M is independent to position of B <br />3. Note that triangle DBC is the image of ABG in the rotational transformation center B , rotational angle= 90 so DC=AG and DC ⊥AG<br />Quadrilateral AMRC is cyclic => ∠ (MAR)= ∠ (MCR)<br />Triangle AMG are congruent to CMD ..( case SAS)<br /> And triangle CMD is the image of AMG in the rotational transformation center M , rotational angle= 90 so MD=MG and MD⊥MG<br />So triangle DMG is an isoceles right triangle<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6474426497981819892017-08-25T16:02:58.471-07:002017-08-25T16:02:58.471-07:00Square off both squares so they are each surrounde...Square off both squares so they are each surrounded by 4 congruent triangles. <br /><br />Let x be the height of E, y be the height of B and z be the height of F. Then one set of triangles has lengths x,y, AB and the other ones z,y,BC.<br /><br />Set the lower left corner to (0,0) for convenience.<br /><br />From there M is at coordinates: y + (x+z)/2, (x+z)/2. Its clear the height is dependent only on x+z i.e. AC likewise the length is as well. As you move C only y can change and its counterbalanced on both sides.<br /><br />This also means if we draw a perpendicular down from M its height is x+z/2 and its located at x+z/2 i.e. the midpoint between AC. So that forms 2 isosceles right triangles that combine to form a larger one ACM. <br /><br />You can compute the coordinates of D and G similarly and show they are inverted slopes away from M: deltaX = y + (z - x)/2 delta y = +-(z - x)/2 - y So DGM is <br />also a right isosceles.<br />Benjamin Leishttps://www.blogger.com/profile/10974191081762367425noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34317500291434806082017-08-25T13:22:41.651-07:002017-08-25T13:22:41.651-07:00Problem 1344
Let EE’,BB’,MM’,FF’ perpendicular at...Problem 1344<br />Let EE’,BB’,MM’,FF’ perpendicular at AC then 1.tri ABB’=tri EE’A=>EE’=AB’,EA=BB’,<br />2.tri BB’C=tri CFF’=>BB’=CF’,B’C=FF’.At the trapezoide EE’F’F MM’ is median so <br />MM’=(EE’+FF’)/2=(AB’+B’C)/2=AC/2 .Is M’ medpoint AC and E’F’ so tri AMC is<br />an isosceles right.In the extention of the B’B to the B point K such that BK=AC.<br />Then tri BKG=tri CAB(S,A,S)(DK=BC=BG=>DKGB=parallelogram.If the DG intersect the KB at H,<br />Then BH=BK/2=AC/2=MM’,(BH//MM’)=>BHMM’= parallelogram =>BM’=HM.<br />Now tri BHG=tri BCM=>BM’=HG=HD=HM=> <LBD+<KDB=90.<br />So the MH is perpendicular at DG.Therefore tri DMG is isosceles and right.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE <br /><br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26719746405248626412009-01-26T09:25:00.000-08:002009-01-26T09:25:00.000-08:00this's actually not hard to prove.T: midpoint ...this's actually not hard to prove.<BR/>T: midpoint of AC<BR/>Let M,N,P be the points so that EM,FN, BP is perpendicular to AC.<BR/>easily have: EM+FN=AC=2MT => MT // EM=> M is independantnatashttps://www.blogger.com/profile/00417040043667077182noreply@blogger.com