Question:

In \( \triangle ABC \), ray \( BD \) bisects \( \angle ABC \), \( A - D - C \), and \( \text{seg } DE \parallel \text{side } BC \). If \( A - E - B \), then for showing \( \frac{AB}{BC} = \frac{AE}{EB} \), complete the following activity: 

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For proving ratios in triangles with parallel lines, use the basic proportionality theorem or the angle bisector theorem effectively.
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Solution and Explanation

Step 1: In \( \triangle ABC \), ray \( BD \) bisects \( \angle B \), so: \[ \frac{AB}{BC} = \frac{AD}{DC} \quad \text{(I)}. \] Step 2: In \( \triangle ABC \), \( DE \parallel BC \), so: \[ \frac{AE}{EB} = \frac{AD}{DC} \quad \text{(II)}. \] Step 3: From equations (I) and (II), we have: \[ \frac{AB}{BC} = \frac{AE}{EB}. \] Hence proved.
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