Prove that GDE GBA using the given condition DE AB and explain the steps.

Step 1: Use the property of similar triangles formed by parallel lines $DE \parallel AB$.
Consider $\triangle GDE$ and $\triangle GBA$.
Since $DE \parallel AB$, we have:
$\angle G$ is common to both triangles.
$\angle GDE = \angle GBA$ (corresponding angles, as $DE \parallel AB$ and $GB$ is a transversal).
$\angle GED = \angle GAB$ (corresponding angles, as $DE \parallel AB$ and $GA$ is a transversal).
Therefore, $\triangle GDE \sim \triangle GBA$ by AAA similarity.
From the similarity, the ratio of corresponding sides is equal:

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