because that reference:In figure ABCD is a parallelogram and also the triangle APB, AQD and also CRD are equilateral. Uncover x. (answer: $120^circ$)


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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