tag:blogger.com,1999:blog-6933544261975483399.post3664437036952458276..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1197: Four Squares, Four Triangles, Equal AreasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-19682031895710181662016-03-07T09:08:37.276-08:002016-03-07T09:08:37.276-08:00Draw altitudes, GG' on BC, JJ' on CD, LL&#...Draw altitudes, GG' on BC, JJ' on CD, LL' on AD, EE' on AB<br />Right tr GCG'=JCJ' (perpend sides) =>GG'=JJ'=>A2=A3<br />Tr JDJ'=LDL' => A3=A4<br />Draw E'E"=CG' =>EBE"=BCG=AEB =>A1=A2c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81862428056775347532016-03-07T02:55:04.855-08:002016-03-07T02:55:04.855-08:00In the 3rd paragraph I should have said
"If ...In the 3rd paragraph I should have said<br /><br />"If we want to avoid trigonometry" Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62375123793430676482016-03-07T00:45:56.397-08:002016-03-07T00:45:56.397-08:00Using the trigonometric formula for the area of an...Using the trigonometric formula for the area of any triangle XYZ, S(XYZ)= XY.YZ sin < XYZ and using another well known trigonometric identity sin@=sin(180-@)we have<br /><br />A1 = BE.BA sin<ABE = BG.BC sin<CBG = A2 = BC.CG sin<BCG = CD.CJ sin <DCJ = A3 = CD.DJ sin CDJ = AD.DC sin ADC = A4<br /><br />If we want to avoid geometry, we can drop perpendiculars from C to BG and A to BE extended and prove thro' congruence ASA that these perpendiculars are equal in length and hence A1 = A2. <br /><br />Similarly by dropping perpendiculars G to BC and J to CD, A2 = A3 and by dropping perpendiculars from C to DJ and L to AD, A3 = A4<br /><br />Hence by trigonometry or geometry A1 = A2 = A3 = A4<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka<br />Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50337489712615299292016-03-06T23:28:14.217-08:002016-03-06T23:28:14.217-08:00Is angle GCB+ angle JCD=180 therefore A2=A3 (BC=CD...Is angle GCB+ angle JCD=180 therefore A2=A3 (BC=CD, CG=CJ) and A3=A4 (CD=DA,DJ=DL, angle CDJ+angleADL=180)and A1=A2(BE=BG BA=BC angleEBA+angleGBC=1800.APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13282000139194507892016-03-06T20:34:02.644-08:002016-03-06T20:34:02.644-08:00Note that ∡ (EBA) supplement to ∡ ( GBC) => sin...Note that ∡ (EBA) supplement to ∡ ( GBC) => sin (EBA)=sin(GBC)<br />A1= ½.EB.BA. sin (EBA)=1/2. BC.BG. sin(GBC)= A2<br />Similarly ∡ (BCG) supplement to ∡ ( DCJ) => sin (BCG)=sin(DCJ)<br />And A2=A3=1/2.BC.CG.sin(BCG)<br />Similarly ∡ (CDJ) supplement to ∡ ( ADL) => sin (CDJ)=sin(ADL)<br />And A3=A4=1/2.DC.DJ.sin(CDJ)<br />So A1=A2=A3=A4<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com