My progress:I found the complying with links and also put it in the drawing$AF = FC\BF = FD\measuredangle QEG = 60^o\measuredangle DAC = measuredangle ACB$
but ns can"t finish..
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Please keep in mind that $angle PAQ = angle BAD$, $AP = AB$, $AQ = AD$.
So $ riangle PAQ cong riangle poor $ and also hence $angle AQP = angle ADB$
Similarly show that $ riangle RDQ cong riangle CDA$ and also so, $angle DQR = angle CAD$
But $angle ADB + angle CAD = 180^circ - x $ (see $ riangle AOD$ whereby $O$ is the allude of intersection that diagonals in the parallel $ABCD$)
That leader to,
$angle PQR = x = 180^circ - x + 60^circ implies x = 120^circ$
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