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 763.
AG/GC = AE/DC = AE/AB = CF/BC = CH/AH1 + AG/GC = 1 + CH/AHAC/GC = AC/AHGC = AHGC - GH = AH - GHCH = AG
Since EF//AC, so let BE/EA = BF/FC = k. Then, FH/HD = EG/GD = 1/(k+1). Let AD=BC=a, AB=CD=b, thenS(ADE) = 1/2*a*b*sin A*1/(k+1)S(CDF) = 1/2*a*b*sin C*1/(k+1)then S(ADE)=S(CDF) (notice that A=C). Therefore, S(ADG)= S(ADE)*DG/DE= S(ADE)*(k+1)/(k+2)= S(CDF)*(k+1)/(k+2)= S(CDF)*DH/DF= S(CDH)Since ADG and CDH have the same height, thus, AG = CH.
Denote the area of triangle PQR by S(PQR). From EF//AC, so BE/EA = BF/FC and EG/GD = FH/HD. AG/CH= S(ADG)/S(CDH)= S(ADE)/S(CDF)= S(ADB)/S(CDB)= 1Hence, AG=CH.
Using the vector approach to solve the problem.Denote vector DA = a, vector DC = c.Let AE:EB=k:1-k, then BF:FC=k:1-k (as triangle BEF ~ triangle BAC)By considering that AG is along AC , AG = rAC = -ra + rcBy considering that AG divides DE into s:1-s, AG = -sa + k(1-s)cHence r = s = k/(1+k)=> AG = (k/(1+k))(c-a)Symmetrically, CH = -(k/(1+k))(c-a)Q.E.D.
Let BD intersect EF, AC, at L and O respectively.In ∆ABC,O is the midpoint of AC and EF ∥ AC imply L is the midpoint of EF.In ∆DEF,L is the midpoint of EF and GH ∥ EF imply O is the midpoint of GH.HenceAG = OA - OG = OC - OH = CH