Given: $ABCD$ is a trapezium with $\text{AB || CD}$
Construction: Draw $DE$ and $CF \perp $ to $AB$
Then in $\triangle\text{ABC}$
$\angle\text{BAC}$ is acute
$\therefore \mathrm{BC}^2=\mathrm{AC}^2+\mathrm{AB}^2-2 \mathrm{AF}: \mathrm{AB} \ldots(1)$
and In $\triangle\text{BDA}$
$\angle\text{DBA}$ is acute
$\therefore \mathrm{AD}^2=\mathrm{BD}^2+\mathrm{AB}^2-2 \mathrm{BE} \cdot \mathrm{AB} \ldots(2)$
Adding $(1)$ and $(2)$ we get
$B C^2+A D^2=A C^2+B D^2+2 A B^2-2 A F \cdot A B-2 B E \cdot A B$
$\Rightarrow A C^2+B D^2=B C^2+A D^2-2 A B[A B-A F-B E]$
$=B C^2+A D^2-2 A B[A B-(A E+E F)-(B F+E F)]$
$=B C^2+A D^2-2 A B[A B-(A E+E F+B F+E F)]$
$=B C^2+A D^2-2 A B[A B-(A B+C D)](\$ \therefore S E F=D C)$
$=B C^2+A D^2-2 A B[(-C D)]$
$=A D^2+B C^2+2 A B \times C D$
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