已知:AB∥CD,點(diǎn)E在CD上,點(diǎn)G、F在AB上,點(diǎn)H在AB、CD之間,連接FE、EH、HG,∠AGH=∠FED,F(xiàn)E⊥HE,垂足為點(diǎn)E.
(1)如圖1,求證:HG⊥HE;
(2)如圖2,GM平分∠HGB,EM平分∠HED,GM、EM交于點(diǎn)M,求證:∠GHE=2∠GME;
(3)如圖3,在(2)的條件下,F(xiàn)K平分∠AFE交CD于點(diǎn)K,若∠KFE:∠MGH=13:5,F(xiàn)K與ME所在直線相交于點(diǎn)Q,求∠FQM的度數(shù).