tag:blogger.com,1999:blog-6933544261975483399.post8074653787188285547..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1329: Two Squares Side by Side, Parallel, Perpendicular, 90 Degrees, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-36104043580787610482023-09-08T02:45:40.416-07:002023-09-08T02:45:40.416-07:00Let the larger square be of side a and the smaller...Let the larger square be of side a and the smaller square be of side b<br />Complete Rectangle CEFN<br /><br />Triangle HNG is Right Isosceles so HN = a<br />Triangle CNM is Right Isosceles so NM = b<br /><br />Further BH = (a+b) - a = b<br /><br />So Triangles AGH & GNH are congruent SAS (AB = HN = a, BH = NM = b & the included angle is 90)<br /><br />Hence HM = AH<br /><br />Also < AHM = 90 since < AHG & < MHN (= < GAH) are complementary<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74269265689049530062017-06-29T10:59:05.674-07:002017-06-29T10:59:05.674-07:00To solve <AHM=90°,
construct point J so that CD...To solve <AHM=90°,<br />construct point J so that CDGJ is a rectangle and BC extended meets J. <br />Triangle ABH is congruent to HMJ <br /><BAH=<JHM<br /><BAH+<BHA+<90°=180° <br /><JHM+<BHA+<AHM=180° <br />Hence <AHM=90°Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89916560840045856712017-06-29T10:50:54.432-07:002017-06-29T10:50:54.432-07:00To prove AH=HM:
Let intersection point of CM and E...To prove AH=HM:<br />Let intersection point of CM and EF be point I. <br /><HGM+<GHM = <HMC+<CMF (ext.<) <br />Since <HGM=45° and <HMC=<GHM<br /><CMF=45° <br />By using this result, it can be deduced that triangle HCE,CEI and IFM are isosceles. <br /><br />Let length of side of large square be 1.<br />BH^2+1^2=AH^2<br />(HC+EF)^2+(CE+FM)^2=HM^2 (both Pyth.)<br /><br />BH+HC=CE+ED=1<br />HC=CE (isos.)<br />So BH=ED<br />HC+ED=1<br />Since ED=EF,<br />HC+EF=1<br /><br />CE=EI (isos.)<br />IF=FM (isos.)<br />EI+IF=EF<br />Hence CE+FM=1<br /><br />ED^2+1^2=AH^2<br />1^2+ ED^2=HM^2 (both identical)<br />Hence AH=HM Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63941841569905995562017-04-13T06:57:34.707-07:002017-04-13T06:57:34.707-07:00Since HCE is right-angled isosceles, => BH=ED
A...Since HCE is right-angled isosceles, => BH=ED<br />Also CM||EG => MG=CE => HCMG is an isosceles trepezoid <br />Join CG and let m(GCD)=x => m(GCM) = 45-x = m(GHM) -----------(1)<br />Drop a perpendicular from H to meet AG at P and since HPG is isosceles triangle => m(PHG) = 45 --------(2)<br />Also the triangles ABH and CDG are congruent (SAS) => m(BHA) = 90-x => m(AHP) = x -------------(3)<br />Therefore m(AHM) = m(AHP)+m(PHG)+m(GHM) = 90 <br />Since m(AGM) = m(AHM) = 90 => AHMG are concyclic<br />Join AM and since m(MGH) = 45 => m(MAH) = 45 and the triangle AHM is right-angled isosceles => AH=HMAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49353031897296134652017-04-10T13:08:34.838-07:002017-04-10T13:08:34.838-07:00https://goo.gl/photos/rZBQU3ty6Rsrpccq6
BC meet F...https://goo.gl/photos/rZBQU3ty6Rsrpccq6<br /><br />BC meet FG at N<br />Observe that HCE and CNM are 45-45-90 triangles<br />So HB=ED=CN=DG<br />Triangles ABH and HNM are congruent => AH=HM and angle AHM= 90<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11869360146464916932017-04-10T10:20:48.612-07:002017-04-10T10:20:48.612-07:00Problem 1329 Solution 2
Suppose that the BC in...Problem 1329 Solution 2<br /> Suppose that the BC intersects GM in point N then NG=NH=AB (<GHN=45=<HGN=<MCN)<br />and CN=NM=DG=BH(AG=BN or AD+DG=BH+HN or DG=BH).Istriangle ABH=triangle HNM<br />so AH=HM and <BAH=<MHN but <BAH+<BHA=90 or <BHA+<MHN=90.Therefore<br /><AHM=90.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41227237461858468092017-04-10T09:47:47.067-07:002017-04-10T09:47:47.067-07:00Problem 1329
Is CM//EG then ECMG=parallelogram...Problem 1329<br />Is CM//EG then ECMG=parallelogram.So CE=MG.But <EHC=<CEH=45=<EGM then <br />HC=CE=MG so HCMG is isosceles trapezoid (cyclic) or HM=CG.But the AHCG is isosceles <br />Trapezoid (cyclic) and so the points A,H,C,M and G are concyclic.Now <HAM=<HGM=45<br />or arc AH=arc HM or AH=HM and <AHM=<AGM=90.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.com