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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view the complete problem 1040.
Using mass point geometry. Let m(P)=mass of point P. Let AB=BC=a. If m(B)=15, then m(B)/5=m(A)/(a-5), so m(A)=3a-15. Also, we have m(B)/3=m(C)/(a-3), so m(C)=5a-15. Hence, m(D)=m(A)+m(B)=3a; m(E)=m(B)+m(C)=5a. Since m(E)=m(D)+m(F), thus m(F)=2a. Now m(D):m(F)=3:2, but also m(D):m(F)=8:x. So 8/x=3/2, and finally x=16/3.
for the theorem of parallels we have 5:3 = (X+8):8 and therefore X= 16/3
Use menelaus theorem on triangle ADF and line BEC and then on triangle ABC and line DEF
EG//CA drawn,with G on AB.`.`BG=BE. .'.GA=BA-BG=BC-BE=EC=3 & so DG=2InΔDAF,GE//AF .'.2/3=x/8 =>x=16/3
by menelaus theorem, triangle BDE and line ACF(DA/AB)(BC/CE)(EF/FD) = 1(5/AB)(BC/3)(8/(8+x)) = 1but AB = BC 40=3(8+x) --> x = 16/3
2 simple ways of doing itDraw a parellel to AC thro D then 2/3 = x/8 Or draw altitudes h1 and h2 from D and E and from similar Tr.s agIn (x+8)/8 = h1/h2 = 5/3From both approaches x = 16/3Sumith PeirisMoratuwaSri Lanka
2 more simple waysGet similar Tr.s by either drawing a parellel to BC thro D or a parellel to AB thro E