tag:blogger.com,1999:blog-6933544261975483399.post3797239874597013964..comments2024-03-17T11:34:28.946-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1334. Square, Kite, Sum of SegmentsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-6933544261975483399.post-13101307914521686602021-04-28T18:35:49.439-07:002021-04-28T18:35:49.439-07:00Read @ as Alpha
Let m(CBE)=m(GBE)=@
It can be obse...Read @ as Alpha<br />Let m(CBE)=m(GBE)=@<br />It can be observed that BGH is isosceles=> BG=c<br />Mark O on GB such that GO=a <br />It can be derived that m(BAO)=@. Extend AO to meet BC at P and BE at Q<br />Since m(BAQ)=@ and m(ABQ)=90-@=>BQ _|_ AQ and hence BOP is isosceles => BP=BO<br />Tr. ABP congruent to BCE (ASA) => BP=CE=b => BO=b<br />GO+BO=a+b=cSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69858459708696296062018-12-26T09:02:32.236-08:002018-12-26T09:02:32.236-08:00EC/BC=DE/DH, EC=b, BC=x, DE=x-b, DH=c-(x-a)
EC/BC=DE/DH, EC=b, BC=x, DE=x-b, DH=c-(x-a)<br />c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85718848973337472472018-12-25T23:41:21.477-08:002018-12-25T23:41:21.477-08:00Can you please explain the first line?Can you please explain the first line?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-24132357770017075902018-05-20T22:09:01.703-07:002018-05-20T22:09:01.703-07:00Excellent solutionExcellent solutionKAMAL SINGHhttps://www.blogger.com/profile/15566453526398693769noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-24135050466801585292017-06-21T04:21:25.254-07:002017-06-21T04:21:25.254-07:00Let length of square be 1.
<CBE=<EHD
<C...Let length of square be 1. <br /><br /><CBE=<EHD<br /><CBE= tan(-1)b<br /><EHD= tan(-1){1÷(a+c)}<br />So, b(a+c)=1<br /><br />BG=GH<br />SQRT(1+a^2)=c (∆BAG)<br />By substituting b(a+c)=1 into SQRT(1+a^2)=c, you get a+b=c.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-22609408206192150022017-05-01T17:49:02.694-07:002017-05-01T17:49:02.694-07:00Triangle GBH is Isosceles and GB=a+b (One way to d...Triangle GBH is Isosceles and GB=a+b (One way to derive is considering the right triangle BFE and applying pythogorus)<br />=> c=a+bAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45686534586736085672017-05-01T13:07:45.252-07:002017-05-01T13:07:45.252-07:00Draw BK ⊥ BE, K on DA. Draw circle through H, B, K...Draw BK ⊥ BE, K on DA. Draw circle through H, B, K<br /> G center of the cirlec.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47803479141929329412017-04-30T21:10:28.245-07:002017-04-30T21:10:28.245-07:00Further result < AFC = 135Further result < AFC = 135Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84902826599815246402017-04-30T21:03:31.499-07:002017-04-30T21:03:31.499-07:00Solution 2
Let AB = p
From similar triangles b/p...Solution 2<br /><br />Let AB = p<br /><br />From similar triangles b/p = p/(a+c) <br />So p^2 = b(a+c) = c^2 - a^2 and hence dividing by a+c, b = c-a<br /><br />Sumith Peiris<br />Moratuwa <br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69376518839884084832017-04-30T20:59:07.554-07:002017-04-30T20:59:07.554-07:00Extend DA to X such that AX = b.
Then Tr.s BCE,BE...Extend DA to X such that AX = b. <br />Then Tr.s BCE,BEF & AXB are all congruent SAS and so < ABX = < GBH = < CBE < GHB.<br /><br />Since AB is perpendicular to XH, it follows that < XBH = 90 <br /><br />But GB = GH so each of these = XG<br />Therefore a+b =c<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4964231272685847942017-04-30T15:39:22.127-07:002017-04-30T15:39:22.127-07:00∆EDH ~ ∆BCE => x²= ab + bc ( x = AD )
From ∆...∆EDH ~ ∆BCE => x²= ab + bc ( x = AD )<br />From ∆ABG => x² = c² - a²<br />From two rows c = a + bc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76083264067668040872017-04-30T12:47:06.464-07:002017-04-30T12:47:06.464-07:00Problem 1334
Is <GBE=<EBC=<EHG then ...Problem 1334<br />Is <GBE=<EBC=<EHG then GH=BG. Draw BK perpendicular in BE at point B (K belongs to ΑD).But triangle KBA=triangle BCE (rectangular, AB=BC,<KBA=<EBC) so KA=CE.Now rectangular triangle KBH is <KBG+<GBH=90=<BKH+<BHK so <KBG=<BKG ,then KG=BG=GH. Therefore GH=KG=KA+AG=CF+AG.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.com