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Geometry (circles and triangles) proofs

  1. Apr 15, 2006 #1
    I'm having some trouble with one particular geometry proof:

    From that I've drawn the following:

    [​IMG]

    [tex]\angle ADB = \angle CED[/tex] (as [tex]\angle ADB[/tex] and [tex]\angle CED[/tex] are alternant sements)



    [tex]\angle CBD = 180 - \angle CED [/tex] (1) (as they are opposite angles in a cyclic quadrilateral)

    [tex]180 - \angle CBD = \angle DBA[/tex] (2) as CBA is a straight line

    [tex]\therefore \angle CED = \angle DBA[/tex] substitute (1) into (2)


    This indicates that [tex]\angle BDA = \angle DBA[/tex]. Is that right? And is that enough to prove that [tex]\bigtriangleup ABD \sim \bigtriangleup CDE[/tex] ?
     
    Last edited: Apr 15, 2006
  2. jcsd
  3. Apr 15, 2006 #2

    VietDao29

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    Nah, this is wrong.
    [tex]\angle ADB \neq \angle CED[/tex]
    It should read:
    [tex]\angle ADB = \angle ECD[/tex]
    (notice that the 2 chords DE = DB).
    Can you see why? :)
     
  4. Apr 16, 2006 #3
    Is it because [tex]\angle ECB = \angle CDA = 90[/tex] which would make EC and DB parrelel in turn making [tex]\angle ADB[/tex] and [tex]\angle ECD[/tex] alternate angles?

    I don't know how to prove this, but if [tex]\angle CDA = 90[/tex] that means CD is the diameter. No that has to be true as the radius (of which the diameter is an extension) meets any tangent at right angles, therefore the converse must be true.

    If CD is the diameter could someone set me on the right track as to how to prove it?
     
  5. Apr 18, 2006 #4

    VietDao29

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    Nah, it's wrong, since the problem does not state that EC and DB are parallel. You cannot assume that...
    --------------
    You should note that:
    [tex]\angle ADB = \angle BCD[/tex] (Tangent chord property)
    [tex]\angle BCD = \angle ECD[/tex] (DE = DB)
    Can you go from here? :)
     
  6. Apr 18, 2006 #5
    One more little hint if you need it...There is a segment on that circle that subtends an angle that you can find fairly easily...it also subtends a different angle :wink:

    *edit* additionally, why is it assumed that the diameter of the circle connects the points D and C??...if you put C somewhere else you still get a secant ABC.
     
    Last edited: Apr 18, 2006
  7. Apr 19, 2006 #6

    Curious3141

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    Who assumed that CD is a diameter ? It isn't in general.
     
  8. Apr 19, 2006 #7
    That it isn't a diameter was also my thought...hence my reason for asking why it was assumed
     
    Last edited: Apr 19, 2006
  9. Apr 19, 2006 #8

    Curious3141

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    Ah, I see. The orig. poster (Beam Me Down) was the only one who thought that (it's not correct).

    I was actually unfamiliar with the tangent chord property that VietDao brought up (till he mentioned it and I googled for it).

    The way I did this was in observing that

    angle DEC = 180 - angle DBC (cyc quad) = angle ABD (supplementary angles).

    Let's extend AD further and call a point on it F (A, D and F are collinear and D lies in between A and F). It should be obvious from the symmetry of DE = DB that angle EDF = angle BDA = theta (say). Then angle BDE = 180 - 2*theta. Angle BCE = 2*theta (opp angles, cyclic quad).

    Now observe that angle DCE = angle DCB (equal chords ED and DB subtend equal angles). Therefore angle DCE = theta = angle BDA.

    Since we've proven two corresponding angle pairs are equal (DEC = ABD & DCE = BDA), we're done.

    In the process of doing this, one can also prove the tangent chord property. :smile:
     
    Last edited: Apr 19, 2006
  10. Apr 20, 2006 #9
    I thought for a second that that I'd proved that [tex]\angle ADC[/tex] was a right angle. But its definately not in all cases.
     
  11. Apr 20, 2006 #10

    Curious3141

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    You see how to do the problem now ? :smile:
     
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