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In the circle of diameter $AB$ it is well known each point $C$ determines a right triangle $\Delta ABC$ and so it is with every point $D$ on the circle of diameter $AC$ determining a right triangle $\Delta ACD$.

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1) Are there points $C$ and $D$ such that $(AB, BC, CA)$ and $(AC, CD, DA)$ be pythagorean triples and the triangles $\Delta ABC$ and $\Delta ACD$ being of equal area?.

2) Assuming $BC = CD$, prove that the possibility that both $(AB, BC, CA)$ and $(AC, CD, DA)$ are pythagorean triples depends on the parity of $BC$, possible in one case and impossible in the other.

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  • $\begingroup$ Where is this from? What have you tried? $\endgroup$ – Pedro M. Jun 10 '15 at 18:41
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    $\begingroup$ 1) is by a desire to know; 2) is by studying concerned diophantine equations. $\endgroup$ – Piquito Jun 10 '15 at 19:58

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