Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1273.
https://goo.gl/photos/bYZ9Rdpg2VBXLXcB8Using pole and polar properties:Let DE and DM cut circle O at F and G,Let MA cut circle O at HWe have OM. OA= OC^2=OD^2 So A is the pole of BC w.r.t circle OAnd A and E are harmonic conjugate points of D and F or ( A,E, D,F)= -1Since ∠EMA= 90 so Apollonius circle diameter AE will pass through MSo ME is a angle bisector of angle MDF.And ∠AMD=∠FMH=∠HMG => BC//GFSo Arc BG=Arc CF and ∠BDM=∠EDC
https://goo.gl/photos/bYZ9Rdpg2VBXLXcB8Using elementary geometry:Let DE and DM cut circle O at F and G,Let MA cut circle O at HWe have OM. OA= OC^2=OD^2So OA/OF=OF/OM => triangle MOF similar to triangle FOA ( case SAS)And FA/FO=FM/MO….. (1)Similarly triangle MOD similar to triangle DOA ( case SAS)And DA/DO=MD/MO ….. (2)Divide (1) to (2) and note that OD=OF We have FA/DA=FM/MD => MA is a external angle bisector of angle DMF And ∠AMD=∠FMH=∠HMG => BC//GFSo Arc BG= Arc CF and angle BDM= angle CDE
Problem 1273If AD and DM cut circle O at K and L.Then AB^2=AM.AO=AC^2=AD.AK. So MOKD is cyclic.OD^2=OB^2=OM.OA so triangle ADO is similar with triangle DMO.Therefore <OMD=<ODAor <AMD=<ODK=<OMK. So <KMC=<CMD =<KMD/2=<KOD/2=<KBD, then <CMD=<CBD+<BDL=<KMC=<CBD+<CBK. So <BDL=<KBC=<KDC.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
Reading thro the 2nd proof of Peter, give below a slightly modified approachLet AD meet the circle at P. Let < MAP = p, < PAC = q, < ACD = < DBC = < DPC = r and < ODP = < OPD = sThen OC^2 = OM. OA = OD^2 = OP^2So < MDO = < MPA = p hence OMDP is concyclic Therefore < AMD = s and so < BDM = 90-s-r .....(1)Now < DOC = 2r so < CDE = 90-r-s (since < ODP = s)Therefore < BDM = < CDE = 90-s-rSumith PeirisMoratuwaSri Lanka
How about the converse "if < BDM = < CDE then M is the mid point"?
The symmedian passes through the intersection of the tangents to the circumcircle at the extremities of the opposite side, i.e. AD is symmedian of tr. DBC.
Extend AD to meet the circle O at X.Connect AMO and form the right triangle ABOIn the triangle ABOAB^2 = AM.AO But AB^2 = AD.AXHence D,M,O and X are concyclic points. Let Angle XDC = EDC = @ and ODX = $=> Angle OXD = $ => OMD = 180-$ => AMD = $ => BMD = 90+$ -------------(1)Join OC and consider the isosceles triangle ODCWe know Angle ODC = @+$ = OCD => Angle DOC = 180-2(@+$) => Angle DBC = DBM = DOC/2 = 90-(@+$)----------(2)Now consider the triangle BDM and from (1) and (2)Angle BDM = 180-(90+$)-(90-(@+$)) = @ = EDC