Thursday, July 8, 2010

Problem 469: Isosceles triangle, Midpoint, Transversal, Congruence

Geometry Problem
Click the figure below to see the complete problem 469 about Isosceles triangle, Midpoint, Transversal, Congruence.

Problem 469: Isosceles triangle, Midpoint, Transversal, Congruence
See also:
Complete Problem 469

Level: High School, SAT Prep, College geometry

5 comments:

  1. Consider triangle FBD with ACE as the transversal. By Menelaus's Theorem, (AB/AD)*(DE/DF)*(CF/BC)=1, ignoring the minus sign. Given DE=DF and AB=BC, we get CF/AD=1 or AD=CF
    Ajit

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  2. Draw perpendicular lines DG and FH to line AC.
    Two right triangle ADG and CFH are congruency, since DG = FH (since DE = EF),
    and angle DAG = angle BCA = angle FCH.
    So, we get AD=CF.

    ReplyDelete
  3. Consider triangle FDB &transversal ACE

    1.therefore by Menelau's thm.
    AD/AB * BC/CF * FE/ED = 1

    2.FE=DE.........(E midpoint)
    therefore FE/DE=1

    3.AD * 1/CF * 1 = 1.....(AB=BC...[given] &
    from 1)

    4.therefore AD=CF.......(rearranging)

    Q.E.D

    ReplyDelete
  4. By Menelaus AB/AD*DE/EF*CF/BC=CF/AD=1 because AB=BC and DE=EF.

    Ivan Bazarov

    ReplyDelete
  5. Draw DG // AC where G is on CB extended

    From mid point theorem CF = CG = CB + BG = AC + BD = AD since Tr. BDG is isoceles

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete