tag:blogger.com,1999:blog-6933544261975483399.post1574396420660398287..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1053: Triangle, Two Perpendicular Medians, Midpoint, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-52321893706614139042015-08-27T11:01:47.118-07:002015-08-27T11:01:47.118-07:00Extend BH to G such that H is mid point of BG. The...Extend BH to G such that H is mid point of BG. Then CH // AG by applying mid point theorem to Tr. ABG. <br />Similarly AH//GC. <br />So AHCG is a //ogram with < AHC = 90, hence the same is a rectangle whose diagonals must be equal.<br />So AC = HG = BH<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2620084471478880092015-01-04T02:36:45.268-08:002015-01-04T02:36:45.268-08:00Fie F simetricul punctului H fata de punctul D=>...Fie F simetricul punctului H fata de punctul D=>BHCF paralelogram si AH=HF(deoarece 2HD=AH,H fiind centrul de greutate al triunghiului ABC)=>CH mediana si inaltime in triunghiul CAF=>CAE isoscel=>AC=CFion radunoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1070110853898863212014-11-06T08:48:27.851-08:002014-11-06T08:48:27.851-08:00Let F be the midpoint of AC
H divides AD as well a...Let F be the midpoint of AC<br />H divides AD as well as CE as 2:1.<br />AD = CE implies AH = CH and HE = HD.<br />So Right Δ’s AHE and CHD are congruent.<br />Thus AE = CD, so BA = BC implying BHF is the perp bisector of AC.<br />Hence ΔAHF is congruent to ΔCHF, <br />and each is right angled isosceles, HF = AF = FC, <br />Hence BH = 2HF = 2 AF = AC.<br />Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60639486817808350732014-11-01T07:48:54.565-07:002014-11-01T07:48:54.565-07:00Obviously H is the centroid of ΔABC.
Thus let M b...Obviously H is the centroid of ΔABC. <br />Thus let M be the mid-point of AC, then BH=2×HM. <br /><br />Now since ΔAHC is right-angled, thus M is the circumcenter. <br />Hence, AC=2×HM=BH. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com