Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 625.
Saturday, June 18, 2011
Problem 625: Triangle, Parallel Lines, Congruence, Angle Bisector
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ReplyDeleteLet 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= x
we 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= 1
So CM, AL and BN concurrent at F per Ceva's theorem and BF is angle bisector of angle ABC
Peter Tran
Denote intersection of
ReplyDeleteAF, 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 P
Again 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 so
CP/PB = CE/AB
Similarly CB//AD implies triangles BRC and ARD are similar and so
BR/RA = BC/AD
Multiplying the last two equations,
(CP/PB)(BR/RA)= BC/AB (noting that CE=AD)
Thus (CR/RF)(FP/PA) = BC/AB
Now denote the areas BFC, CFA, AFB by x, y, z respectively
Clearly
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/QA
Follows CQ/QA = CB/BA
Hence BF(Q) bisects angle ABC