Is the Common Base Theorem Applicable to All Cases Involving Triangles?

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jobsism
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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?
 
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jobsism said:
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?



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
 
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.
 
jobsism said:
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.


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
 
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 learned 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.
 
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