Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 750 details.
http://img19.imageshack.us/img19/138/problem750.pnglet AC cut EF at P ( see sketch)∆DMA similar to ∆DNF so DA/DF=MA/FN=a/f∆BMA similar to ∆BGE so BE/BA=GE/AM= e/aIn triangle ∆AEF, applying Ceva’s theorem => PE/PF x DF/DA x BA/BE=1Replace DA/DF=a/f and BE/BA=e/a we get PE/PF=e/fIn triangle AEF, applying Von Aubel’s theorem => EC/CD=BE/BA+PE/PFReplace PE/PF=e/f and BE/BA=e/a we get EC/CD=e(f+a)/afCD/DE=CD/(CD+EC)=af/(af+ef+ea)=c/ee/c= 1+e/a+e/fdivide both side by e we get 1/c=1/e+1/a+1/f
Draw a line through B parallel to CH to meet EC at K and DA extended at K’.Let BK = b and BK’ = b’Draw a line through D parallel to CH to meet CF at K and BA extended at L’.Let DL = d and DL’ = d’Apply result of Problem 746 to ∆BCD to obtain c = bd /(b + d), 1/c = 1/b + 1/d ... (i)Apply result of Problem 746 to ∆BAD to obtain a = b'd' /(b' + d'), 1/a = 1/b' + 1/d'... (ii)From similar ∆s EGB and LDB we get e/d' = BG/BD = GD/BD - 1 = (e/b) - 1, 1/d' = 1/b - 1/e ... (iii)Exactly in the same manner, we get 1/b' = 1/d - 1/f ... (iv)Thus by (i) through (iv)1/a = (1/d - 1/f) + (1/b - 1/e) = (1/b + 1/d) - 1/e - 1/f = 1/c - 1/e - 1/fHence 1/c = 1/a + 1/e + 1/f as required
Here is a much simpler Proof:c/e = DH/DG, c/f = BH/BNSo BD/c = (BH + DH)/c = BH/c + DH/c = BN/f + DG/eNext a/e = BM/BG, a/f = MD/DNSo BD/a = (BM + MD)/a = BM/a + DM/a = BG/e + DN/fThereforeBD/c - BD/a = (DG - BG)/e + (BN - DN)/f = BD/f + BD/eFollows 1/c - 1/a = 1/f + 1/eHence 1/c = 1/a + 1/e + 1/f