两道八下几何证明题!不是很难,只需要列算式!

1,（1）设CE交BD与F,交AD于G,
∠B+∠E+∠BFE
=∠B+∠E+∠D+∠CGD
=∠B+∠E+∠D+∠A+∠C=180
(2)∠CBD+∠DBE+∠ABE
=∠A+∠D+∠DBE+∠C+∠E=180
2,设BD交CF于N,交CE于M;CF交AG于X,交AD于H
∠B+∠E+∠BME
=∠B+∠E+∠C+CNM
=∠B+∠E+∠C+∠D+∠CHD
=∠B+∠E+∠C+∠D+∠A+∠AXH
=∠B+∠E+∠C+∠D+∠A+∠G+∠F=180