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如圖取M為AB中點(diǎn), MG交AE于N
∵EC=CD ∠1= ∠2 ∠ C= ∠C
∴△ECG≌△FCD
∴CE=FC FC=1/2EC=1/2CD
∴CG=1/2CD∴G為CD中點(diǎn)
∴MG∥AD∥BC
∴ ∠GEC= ∠EGN (這是關(guān)鍵條件后面會(huì)直接用)標(biāo)記為1
∵AE⊥BC
∴AE⊥CN N為AE中點(diǎn)(這兒由MN是△ABE中位線可得)
由此可得△ANG≌△ENG
∴ ∠EGN= ∠NGA 加上1處
得 ∠CEG=1/2 ∠AGE