tag:blogger.com,1999:blog-6933544261975483399.post7560232358082308691..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 242: Triangle with Equilateral triangles, ParallelogramAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-91401543301119445392020-11-02T09:05:38.999-08:002020-11-02T09:05:38.999-08:00Tr. ABC is congruent to AB'C' and also to ...Tr. ABC is congruent to AB'C' and also to A'B'C, SAS in each case.<br /><br />So B'C' = BC = A'B and also A'B' = AB = BC'<br /><br />So the opposite sides of quadrilateral A'BC'B' are equal whence the same is a parallelogram<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38229938912643350732020-10-08T09:16:52.598-07:002020-10-08T09:16:52.598-07:00See the drawing
- ΔACB’ and ΔBCA’ are equilateral...See the <a href="http://sciences.heptic.fr/2020/10/08/gogeometry-problem-242/" rel="nofollow"><b>drawing</b></a><br /><br />- ΔACB’ and ΔBCA’ are equilateral => ∠ACB’ =∠BCA’ = Π/3<br />- ∠ACB = ∠ACB’ - ∠B’CB = Π/3 - ∠B’CB <br />- ∠B’CA’ = ∠BCA’ - ∠B’CB = Π/3 - ∠B’CB <br />- =>∠ACB =∠B’CA’ <br />- ΔACB is congruent to ΔB’CA’ (SAS) => BA=B’A’<br />In the same way:<br />- ΔCAB’ and ΔBAC’ are equilateral => ∠CAB’ =∠BAC’ = Π/3<br />- ∠CAB = ∠CAB’ - ∠BAB’ = Π/3 - ∠BAB’ <br />- ∠B’AC’ = ∠BAC’ - ∠BAB’ = Π/3 - ∠BAB’ <br />- =>∠CAB =∠B’AC’ <br /><br />- ΔCAB is congruent to ΔB’AC’ (SAS) => BC=B’C’<br />- BA=B’A’ and BA=BC’ (ΔABC’ equilateral) => B’A’=BC’<br />- BC=B’C’ and BC=BA’ (ΔBCA’ equilateral) => B’C’=BA’<br />- => B’A’=BC’ and B’C’=BA’ <br />- <b>=> B’A’BC’ is a parallelogram</b><br />rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71027744040271000722012-11-26T14:45:26.552-08:002012-11-26T14:45:26.552-08:00http://img15.imageshack.us/img15/857/problem242a.p...http://img15.imageshack.us/img15/857/problem242a.pngAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74103209563754112152009-06-16T22:05:00.866-07:002009-06-16T22:05:00.866-07:00C'A = BA
angle C'AB' = Angle BAC = 60...C'A = BA <br />angle C'AB' = Angle BAC = 60 - angle B'AB<br />AB' = AC<br />So, by SAS postulate,<br />triangle C'AB' is congruent to triangle BAC<br />So, C'B' = BC = BA'<br />Similarly as <br />triangle A'B'C is congruent to triangle BAC,<br />A'B' = BA = BC'<br /><br />So,<br />in quadrilateral A'BC'B' both pairs of opposite sides are congruent.<br />So, it is a parallelogram.Mrudul M. Thattenoreply@blogger.com