Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 775.
Friday, June 29, 2012
Problem 775: Triangle, Isogonal lines, Congruent angles, Perpendicular, Metric Relations
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Angle EBM = Angle DBF
ReplyDeleteSo Rt triangles EMB and DFB are similar
∴ x : 8 = BE : BD
Next angle HBE = angle GBD
So Rt triangles HBE and GBD are similar
∴ BE : BD = EH : DG = 25 : 28
Follows x : 8 = 25 : 28
Hence x = 50/7
Triangle BEM and BFD are similar (AA)
ReplyDeleteTriangle BDG and BHE are similar (AA)
Hence DG/EH =BD/BE =ED/EM
28/25 =8/EM
So EM =50/7
Triangle BDG similar to ∆ BEH ….. ( Case AA)
ReplyDeleteSo BD/BE= 28/25
Triangle BFD similar to ∆ BME ….. ( case AA)
So DF/EM=8/x=BD/BE=28/25
So 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=50/7
ReplyDeleteGreetings go-solvers.
x/8 = BD/BE = 28/25 from which x = 200/28 = 50/7
ReplyDelete