Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a trapezium ABCD, AB is parallel to CD and AB$<$CD. Given that AC+BC=AD+BD, prove that AD=BC.

I have a feeling that a proof by contradiction is possible, but to me the result just seems obvious and I have been going round and round in circles trying to find a rigorous proof. Any suggestions?

share|improve this question
    
prove that it is an parallelogram –  Willemien Aug 23 '13 at 10:37
    
how would that be possible? –  heron1000 Aug 23 '13 at 10:44
    
because it is a parallelogram (no other parallel trapezium has AC+BC=AD+BD ) so start that is is not a parallelogram get into a contradiction and you have proved it is a parallelogram, and the rest is simple –  Willemien Aug 23 '13 at 12:55

1 Answer 1

Hint:

  • Consider an ellipse $\mathcal{E}$ with foci at $A$ and $B$, and passing through point $C$.
  • From properties of ellipse we know that for any point $X \in \mathcal{E}$ we have $AX + BX = AC + BC$, in particular $D \in \mathcal{E}$.
  • Finally, ellipses are symmetrical along their axes.

I hope this helps $\ddot\smile$

share|improve this answer
    
Thanks, this really helped. However, I was wondering if a purely synthetic solution is possible, especially since the problem itself looks rather simple. –  heron1000 Aug 23 '13 at 10:44
    
Which part of this approach is not synthetic enough for you? –  dtldarek Aug 23 '13 at 11:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.