1.
如图,\(EF/\!/AD\),\(∠1=∠2\),\(∠BAC=70^{\circ}.\)求\(∠AGD\)的度数\(.\)
解:因为\(EF/\!/AD\), 所以\(∠2=\)____\( (\)_________________________________\()\)
又因为\(∠1=∠2\)
所以\(∠1=∠3\) \((\)__________________\()\)
所以\(AB/\!/\)_____\( (\)___________________________________\()\)
所以\(∠BAC+\)______\(=180^{\circ}(\)___________________________\()\)
因为\(∠BAC=70^{\circ}\)
所以\(∠AGD=\)_______