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Common base theorem doubt

  1. Apr 11, 2012 #1
    Consider the case of a rectangle ABCD, with diagonals AC and BD drawn.

    Now, it's easy to see that triangles ADC and BCD are congruent. So, the ratio of their areas would be 1.

    But when I try to obtain the same result via the common base theorem, the line passing through A and B never meets the base DC (as AB and DC are parallel!). How is this possible, when the ratio of the areas is known to be finite?! Or does this situation imply that the Common Base theorem can't be applied in such cases?
     
  2. jcsd
  3. Apr 11, 2012 #2


    I think it'd be a good idea that you'd tell us what's that "common base theorem" that seems to be bugging you...

    DonAntonio
     
  4. Apr 11, 2012 #3
    Oh, sorry about that. I just thought it was quite a popular theorem. :D

    If two triangles ABC and A'BC have a common base BC, and the line passing through A and A' meets the base BC(extended, if needed) at P, then

    Area(Triangle ABC)/Area(Triangle A'BC) = AP/A'P
    This is the Common Base Theorem.

    But you see, in the above case the AP and A'P counterparts simply extends to infinity, whereas the ratio of the areas is finite.
     
  5. Apr 11, 2012 #4

    Never heard of such a theorem, and it even looks slightly suspicious to me, but it never matters: as you wrote, "...IF the line thorugh A, A' meets the base BC..." , and in the rectangle's case you described it does NOT meet the base, so the theorem isn't appliable.

    DonAntonio
     
  6. Apr 11, 2012 #5
    EDIT: I totally understand your reason for suspicion; it's not mentioned anywhere on the web! I think it might be a lesser known lemma. But the book that I learnt it from (A Primer for Mathematical competitions) called it CBT!
    Anyways, the proof is really simple: just constructing altitudes and using similarity criteria.

    Oh,wow...so silly of me! It was all in the statement the whole time! :D

    Funny you should find the theorem suspicious though. I believe it's used in the proof of the very famous Ceva's theorem. Ah, but then again, there are numerous proofs for the latter.
     
    Last edited: Apr 11, 2012
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