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.
connect AE, connect CF, AE//CF;extend EM, extend CF, cross at G, <CAE=<ACF =<ACG, <AME=<GMC, AM=MC triangle AEM = triangleGMC,so GM=ME, triangle EGF = right angle, so EM=MF
https://goo.gl/photos/9ENrkK4G1b2hs1BCALet G and H are the midpoints of AD and DCWe have AM=GH= ½.ACGM= AM-GA=GH-GD= DHSo GE=GD=MHAnd GM=DH=FHTriangles EGD and DHF are isosceles => ∠EGD=∠DHFTriangles GME congruent to HFM ( case SAS)=> ME=MF
Problem 1302Is <AED=90=<DFC then AE//CF or AEEF is trapezoid.Let N midpoint in EF, so MN//AE//CF then MN is perpendicular bisector EF.Therefore ME=MF.APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE
Solution 1Let the circles be P (larger) and Q (smaller) with radii p and q respectively.So AM = p+q and hence PM = q and QM = p.Therefore easily ∆MPE ≡ ∆FQM (SAS) and hence,ME = MFSumith PeirisMoratuwaSri Lanka
Solution 2Draw MN // AE//FC, N on FELet EN = u, ND = v and DF = wFrom similar ∆s,u/v = (p+q)/(p-q) and w/v = 2q/(p-q)Subtracting, u/v – w/v = (p+q-2q)/(p-q) = 1Hence u-w =v and so EN = NF implying that ME = MFSumith PeirisMoratuwaSri Lanka
Solution 3Let FM extended meet AE at L∆MAL ≡ ∆MFC (ASA)So LM = MF = ME since < LEF = 90Sumith PeirisMoratuwaSri Lanka
See that EF is the projection of AC onto BD, wherefrom the solution: the midpoint of the segment is equally apart of the ends of the projection.Best regards
AE and CF are perpendicular to BD, Since M is midpoint of AC it lies on perpendicular bisector of EF. Hence ME=MF