Three by One
$ABCD$ is a rectangle where $BC$ = $3AB$. $P$ and $Q$ are points on $BC$ such that $BP$ = $PQ$ = $QC$.
Show that: angle $DBC$ + angle $DPC$ = angle $DQC$
Generalise this result.
N.B. This problem can be tackled in at least 8 different ways using different mathematics learnt in the last two years in school and earlier. The methods are essentially the same when viewed from a more advanced perspective.
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