∠EFC´=79°・・・①
①ならば∠EFB=101°・・・②
②ならば∠C´FB=22°
四角形ABCDは長方形だから∠ABF=90°
辺ABと辺C´Fの交差点をXとすると、∠C´FB+∠ABFは∠FXBの外角と等しいので22°+90°=112°。なのでX=112です
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