tag:blogger.com,1999:blog-6933544261975483399.post6196675035997839022..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 240: Triangle with Equilateral triangles, ParallelogramAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-18826510567713970952020-10-08T10:05:16.509-07:002020-10-08T10:05:16.509-07:00It is not difficult to see that < A'CB'...It is not difficult to see that < A'CB' = C and hence Tr.s A'B'C & ABC are congruent whence A'B' = BC'<br /><br />By similar reasoning A'B = B'C'<br /><br />In quadrilateral A'B'C'B the opposite sides are thus equal and the result follows.<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43691340626518712772020-10-06T01:00:48.572-07:002020-10-06T01:00:48.572-07:00See the drawing
ΔCAB’ and ΔC’AB are equilateral
=...See the <a href="http://sciences.heptic.fr/2020/10/06/gogeometry-problem-240/" rel="nofollow"><b>drawing</b></a><br /><br />ΔCAB’ and ΔC’AB are equilateral<br />=> ∠CAB’=∠C’AB=Π/3<br />∠BAB’=a+∠CAB’=∠BAC’+a’<br />=> a=a’<br />AC’=AB and AB’=AC and a=a’=> ΔBAC and ΔB’AC’ are congruent (SAS)<br />=>B’C’=BC<br />But BC=BA’=>B’C’=BA’<br />In the same way : ∠BCA’=∠B’CA=Π/3<br />∠BCA=c= Π/3 - ∠ACA’<br />∠B’CA’=c’= Π/3 - ∠ACA’ =>c=c’<br />CB=CA’ and CA=CB’ and c=c’=> ΔBCA and ΔA’CB’ are congruent (SAS)<br />=>AB=A’B’<br />But AB=BC’ => A’B’=BC’<br /><br />B’C’=BA’ and A’B’=BC’ => BA’B’C’ is a parallelogram<br /><br /><br /><br />rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15066098069043716102010-06-22T02:13:14.934-07:002010-06-22T02:13:14.934-07:001. Triangles AA’B’ and ABC are congruent ( AB’=AC...1. Triangles AA’B’ and ABC are congruent ( AB’=AC , AC’=AB, angle B’AC’= angle CAB ) so BC’=BC=BA’<br />2. Triangles CA’B’ and CBA are congruent ( CB’=CA, CA’=CB, angle B’CA’= angle ACB) <br /> so B’A’=BA=BC’<br />3. Quadrilateral BA’B’C’ have opposite sides congruent so BA’B’C’ is a parallelogram.Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53109907817567937752010-02-19T10:47:10.927-08:002010-02-19T10:47:10.927-08:00ang C'AC = 60 - A =>
B'AC' = A
A...ang C'AC = 60 - A =><br />B'AC' = A<br />A'CB' = C<br /><br />extend BA' to P ( P on AB'), BC' to G, ( G on B'C )<br />BPB' = B - 60 + 60 + A, BGB' = B - 60 + 60 + C =><br /><br />BPB' = A + B<br />BGB' = B + C<br /><br />▲ABC and ▲AC'B' are congruent<br />1) AB = AC'<br />2) BAC = C'AB' = A<br />3) AC = AB'<br />=> <br />AC'B' = B, AB'C' = C<br />=><br />BPB' + AB'C' = (A + B) + C = 180°<br />=><br /><br />BP//B'C'<br /><br />BGB' + GB'A' = (B + C) + A = 180°<br />=><br /><br />BC'//A'B'<br />------------------------------------------c .t . e. onoreply@blogger.com