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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Obviously H is the centroid of ΔABC. Thus let M be the mid-point of AC, then BH=2×HM. Now since ΔAHC is right-angled, thus M is the circumcenter. Hence, AC=2×HM=BH.
Let F be the midpoint of ACH divides AD as well as CE as 2:1.AD = CE implies AH = CH and HE = HD.So Right Δ’s AHE and CHD are congruent.Thus AE = CD, so BA = BC implying BHF is the perp bisector of AC.Hence ΔAHF is congruent to ΔCHF, and each is right angled isosceles, HF = AF = FC, Hence BH = 2HF = 2 AF = AC.
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=CF
Extend BH to G such that H is mid point of BG. Then CH // AG by applying mid point theorem to Tr. ABG. Similarly AH//GC. So AHCG is a //ogram with < AHC = 90, hence the same is a rectangle whose diagonals must be equal.So AC = HG = BHSumith PeirisMoratuwaSri Lanka