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Therefore: S△ADB+S△ADC= S△ABC.

Therefore:1/2× ab× de+1/2× AC× df =1/2× ab× CG.

Because AB=AC (both sides are divided by 1/2×AB at the same time)

Therefore: DE+DF=CG.

(2)∣DE-DF∣=CG

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Therefore: ∣S△ADB-S△ADC∣= △ABC.

Therefore: ∣1/2× ab× de-1/2× AC× df ∣ =1/2× ab× CG.

Because AB=AC (both sides are divided by 1/2×AB at the same time)

Therefore: ∣DE-DF∣=CG.

(Note: Absolute value refers to "big" minus "small")

This problem can also be solved by CG passing through the vertical line of D.