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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 775.
Angle EBM = Angle DBFSo Rt triangles EMB and DFB are similar∴ x : 8 = BE : BDNext angle HBE = angle GBDSo Rt triangles HBE and GBD are similar∴ BE : BD = EH : DG = 25 : 28Follows x : 8 = 25 : 28Hence x = 50/7
Triangle BEM and BFD are similar (AA)Triangle BDG and BHE are similar (AA)Hence DG/EH =BD/BE =ED/EM28/25 =8/EMSo EM =50/7
Triangle BDG similar to ∆ BEH ….. ( Case AA)So BD/BE= 28/25Triangle BFD similar to ∆ BME ….. ( case AA)So DF/EM=8/x=BD/BE=28/25So x=50/7
Let be P on BD such that BP=BE, and let be Q and N the perpendiculars from P to AB and BC respectively. Note that BPQ and BME are congruent, and that BPQ is similar with BDF and BNP is similar with BGB (AA). Hence x=50/7Greetings go-solvers.