# Plane Geometry - Proof by Contradiction?

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?

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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

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 $\ddot\smile$

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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