Friday, August 29, 2014

Geometry Problem 1040: Isosceles Triangle, Transversal Line, Metric Relations

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.

Online Math: Geometry Problem 1040: Isosceles Triangle, Transversal Line, Metric Relations.


  1. 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.

  2. for the theorem of parallels we have 5:3 = (X+8):8 and therefore X= 16/3

  3. 8/(x+8)=[ECF]/[DCF]=[EAF]/[DAF]=([EAF]-[ECF])/([DAF]-[DCF])=[EAC]/[DCA]=3/5
    40=3x+24, x=16/3

  4. Use menelaus theorem on triangle ADF and line BEC and then on triangle ABC and line DEF

  5. EG//CA drawn,with G on AB.
    .'.GA=BA-BG=BC-BE=EC=3 & so DG=2
    InΔDAF,GE//AF .'.2/3=x/8 =>x=16/3

  6. by menelaus theorem, triangle BDE and line ACF
    (DA/AB)(BC/CE)(EF/FD) = 1
    (5/AB)(BC/3)(8/(8+x)) = 1
    but AB = BC
    40=3(8+x) --> x = 16/3

  7. 2 simple ways of doing it

    Draw 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 = 5/3

    From both approaches x = 16/3

    Sumith Peiris
    Sri Lanka

  8. 2 more simple ways

    Get similar Tr.s by either drawing a parellel to BC thro D or a parellel to AB thro E