tag:blogger.com,1999:blog-6933544261975483399.post6217956531301078434..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 49Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-40225052764907863452019-08-22T08:59:14.595-07:002019-08-22T08:59:14.595-07:00With the great help of my friend Greg, we have fou...With the great help of my friend Greg, we have found an original solution.<br /> <a title="See the video" href="https://youtu.be/QXunMNKD9xM" rel="nofollow"> See original solution on this video</a><br /><br />ang(BAC) = 90 - 2α => ang(BAD) = 90 - 2α and ang(ABD) = 2α<br />Let FJ parallel to BC, J on BD<br />BJFE is a parallelogram, BJ = EF and ang(DFJ) = 2α<br />ABE right in B and AEF right in F => A, B, E and F are concyclic, the center O of the circle being the middle of AE<br />Let BD intersect C in K<br />ABO isosceles in O => ang(ABO) = α and ang(ABK) = 2α : BO is the angle bisector of ang(ABK) and AB = BK<br />ang(ABK) = ang(AFK) = ang(DFK) = ang(DFJ) = 2α so on BD J and K are mirror images of one another about AC<br />So DJ = DK, 2BD = BJ + JD + BK – DK and since BJ = EF and BK = AB then 2BD = EF + AB => BD = (AB+EF )/2<br />QED.rv.littlemanhttps://www.blogger.com/profile/05572092955468280791noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45328935437082818732018-09-30T11:24:13.790-07:002018-09-30T11:24:13.790-07:00Reflect AE along AB to AK, triangle ABE = ABK;
are...Reflect AE along AB to AK, triangle ABE = ABK;<br />area AEC = area ABC- area AEK; area AEC =1/2*EF*AC;<br />area AKC =area ABC + area ACK =area ABC + area ABE; area AKC=1/2*AB*KC=1/2*AB*AC;<br />area AEC+ area AKC = 2 area ABC; area ABC=1/2*BD*AC;<br />so BD=1/2(AB+EF);<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18564182094688582422016-02-24T23:24:07.243-08:002016-02-24T23:24:07.243-08:00http://s11.postimg.org/6wmej6q37/pro_49.png
Draw B...http://s11.postimg.org/6wmej6q37/pro_49.png<br />Draw BG and GH as per sketch<br />Observe that ∠ (GAB)= alpha and ∠ (GAE)= 2.alpha<br />Triangle GAE similar to tri. GAC … ( case AA)<br />So both tri. GAE and GAC are isosceles<br />In isosceles triangle AGC , heights AB and GH are congruent <br />In trapezoid GHFE we have BD=1/2(GH+EF)= ½(AB+EF) <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85455458506689149932016-02-24T04:05:11.227-08:002016-02-24T04:05:11.227-08:00Let AJ be perpendicular to CB, J on BC so that <...Let AJ be perpendicular to CB, J on BC so that < JAB = 2@. Let AH be the bisector of < JAB, H on BC <br /><br />So < JAH = < HAB = < BAE = @. Now drop a perpendicular from H to AC, HG with G on AC<br /><br />Tr. s AHG and ABE are congruent ASA since Tr.s ABE and ABH are congruent ASA<br /><br />So AB = HG and the result follows since HB = BE<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13313621376342689272012-04-05T15:48:33.137-07:002012-04-05T15:48:33.137-07:00To Newzad
Please let me know what is the language ...To Newzad<br />Please let me know what is the language you write in and the corresponding nationality.<br />(Just curiosity. I'm brazilian and speek portugese).<br />Thanks.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68632811940201140162012-04-05T15:43:08.429-07:002012-04-05T15:43:08.429-07:00Let’a write a for alpha. Take G symmetric of A ove...Let’a write a for alpha. Take G symmetric of A over the angle C’s bisector, and GH perpendicular to AC.<br />We have ang(BAD) = 90º-2a, and ang(GAC) = 90º-a, so ang(BAG) = a. That means that AEG is isosceles, with GB = BE.<br />Thus B is middle point of GE, then GH - BD = BD - EF. But GH = AB by symmetry,<br />so AB – BD = BD – EF and BD = (AB + EF)/2.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19100326000295906742011-01-26T07:43:26.339-08:002011-01-26T07:43:26.339-08:00AB=AE cos alpha and EF=AE cos 3alpha.Then add the ...AB=AE cos alpha and EF=AE cos 3alpha.Then add the two to get the desired result.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57196034536317430922009-03-17T07:15:00.000-07:002009-03-17T07:15:00.000-07:00It's possible to solve this problem without using ...It's possible to solve this problem without using trigonometry.<BR/><B>The key is auxiliary construction</B>:<BR/><BR/>Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2184599900817484522009-03-17T05:35:00.000-07:002009-03-17T05:35:00.000-07:00Let alpha =a. From Tr. ABC, we've Ang EAF = 90...Let alpha =a. From Tr. ABC, we've Ang EAF = 90 - 3a. EF=AEsin(90-3a)=AEcos(3a). But AB/AE=cos(a). Thus EF=ABcos(3a)cos(a). Hence, (AB+EF)/2 =(ABcos(3a)/cos(a)+AB)/2 =(AB/2)(cos(a)+cos(3a)/cos(a)) =(AB/2)(2cos(a)cos(2a))/cos(a)) = ABcos(2a) or (AB+EF)/2 =ABcos(2a)-----(1). Now from Tr. ABD BD/AB =sin(90-2a) =cos(2a) or BD=ABcos(2a)----(2)<BR/>(1) & (2) give us, BD=(AB+EF)/2<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com