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.Details: Click on the figure below.
Problem 1329Is CM//EG then ECMG=parallelogram.So CE=MG.But <EHC=<CEH=45=<EGM then HC=CE=MG so HCMG is isosceles trapezoid (cyclic) or HM=CG.But the AHCG is isosceles Trapezoid (cyclic) and so the points A,H,C,M and G are concyclic.Now <HAM=<HGM=45or arc AH=arc HM or AH=HM and <AHM=<AGM=90.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Problem 1329 Solution 2 Suppose that the BC intersects GM in point N then NG=NH=AB (<GHN=45=<HGN=<MCN)and CN=NM=DG=BH(AG=BN or AD+DG=BH+HN or DG=BH).Istriangle ABH=triangle HNMso AH=HM and <BAH=<MHN but <BAH+<BHA=90 or <BHA+<MHN=90.Therefore<AHM=90.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
https://goo.gl/photos/rZBQU3ty6Rsrpccq6BC meet FG at NObserve that HCE and CNM are 45-45-90 trianglesSo HB=ED=CN=DGTriangles ABH and HNM are congruent => AH=HM and angle AHM= 90
Since HCE is right-angled isosceles, => BH=EDAlso CM||EG => MG=CE => HCMG is an isosceles trepezoid Join CG and let m(GCD)=x => m(GCM) = 45-x = m(GHM) -----------(1)Drop a perpendicular from H to meet AG at P and since HPG is isosceles triangle => m(PHG) = 45 --------(2)Also the triangles ABH and CDG are congruent (SAS) => m(BHA) = 90-x => m(AHP) = x -------------(3)Therefore m(AHM) = m(AHP)+m(PHG)+m(GHM) = 90 Since m(AGM) = m(AHM) = 90 => AHMG are concyclicJoin AM and since m(MGH) = 45 => m(MAH) = 45 and the triangle AHM is right-angled isosceles => AH=HM