Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 625.
http://img31.imageshack.us/img31/8388/problem625.pngLet CD cut AB at M and AE cut BC at L and angle bisector of angle ABC cut AC at N ( see picture)Let AD=CE= xwe have LB/LC= x/c ( tri. LAB similar to Tri. LEC)MB/MA=a/x ( Tri. MAD similar to tri. MBC)NA/NC= c/a ( BN is angle bisector of angle(ABC))Note that LB/LC . NA/NC . MC/MA= x/c .c/a .a/x= 1So CM, AL and BN concurrent at F per Ceva's theorem and BF is angle bisector of angle ABCPeter Tran
Denote intersection of AF, BC by P; BF, AC by Q; CF, AB by R.(AQ/QC)(CP/PB)(BR/RA) = 1 by Ceva.Consider triangle AFC and point B.BF intersects AC at Q; BA intersects CF at R; BC intersects AF at PAgain by Ceva, (AQ/QC)(CR/RF)(FP/PA) = 1 It follows from the above two equations that(CP/PB)(BR/RA)= (CR/RF)(FP/PA) Now AB // CE implies triangles CPE and BPA are similar and soCP/PB = CE/ABSimilarly CB//AD implies triangles BRC and ARD are similar and soBR/RA = BC/ADMultiplying the last two equations,(CP/PB)(BR/RA)= BC/AB (noting that CE=AD)Thus (CR/RF)(FP/PA) = BC/ABNow denote the areas BFC, CFA, AFB by x, y, z respectivelyClearly CR/RF = (BRC) /(BRF)= (ARC)/(ARF) = [(BRC) + (ARC)] /[(BRF) + (ARF)]=(ABC)/(AFB)Similarly,FP/PA = (BFC)/(ABC)Thus (CR/RF)(FP/PA)= (BFC) /(AFB) = CQ/QAFollows CQ/QA = CB/BAHence BF(Q) bisects angle ABC
Call intersection of DC and AB to be X, and intersection of AE and BC to be Y. AX/XB=DA/BC, and BY/YC=AB/CE. By Ceva, if BF cuts AC at point Z, then CZ/ZA*AX/XB*BY/YC=CZ/ZA*DA/BC*AB/CE=CZ/ZA*AB/BC=1, so CZ/ZA=BC/AB. By angle bisector theorem, BF must bisect <ABC.
Let AE, BC cut at X.Let AD = CE = p, AF = m, FX = n, BX = t, CX = s.From similar triangles,p/s = m/n …(1) andp/c = s/t ….(2)From (1) and (2), m/n =c/t and so BF bisects <BSumith PeirisMoratuwaSri Lanka