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Bunuel, MathRevolution

Seems like figure is missing
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=>

Attachment:
10.17ps(a).png
10.17ps(a).png [ 24.26 KiB | Viewed 2291 times ]

We draw lines \(p\) and \(q\) parallel to \(l\) and \(m.\)

When we put \(<DEF = <DHI = <x\) and \(<GDC=<y,\) we have \(<HDJ=<DHC=180°-x\)

Since we have \(<ABG=<GCD\) and \(<GDC=<DCH\), we have \(60°=30°+<y\) and \(<y=30°.\)

Since \(<GDC+<CDH+<HDJ=180°\) or \(<y+90°+180°-<x=180°\), we have \(30°+270°-<x=180°, 300°-<x=180°\) or \(<x = <DEF = 120°.\)

Therefore, C is the answer.
Answer: C
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