Question

16 what is the missing step in this proof? a. ∠abc ≅ ∠dbe b. ∠bca ≅ ∠bde c. ∠acb ≅ ∠deb d. ∠bde ≅ ∠ade e. ∠cab ≅ ∠dac

Explanation:

To prove the proportion \(\frac{AD}{DB}=\frac{CE}{EB}\), we can use the similarity of triangles. If \(\angle ABC\cong\angle DBE\) (common angle) and we assume some parallel lines (by the basic proportionality theorem or Thales' theorem), we can establish similarity of triangles. For example, if \(DE\parallel AC\), then \(\triangle BDE\sim\triangle BAC\) and the ratio of sides can be derived. The other options:

• Option B: \(\angle BCA\) and \(\angle BDE\) are not in a position to directly give the required proportion.
• Option C: \(\angle ACB\) and \(\angle DEB\) are not relevant for the proportion \(\frac{AD}{DB}=\frac{CE}{EB}\) in the context of triangle similarity for this proportion.
• Option D: \(\angle BDE\) and \(\angle ADE\) are adjacent angles and not useful for the proportion.
• Option E: \(\angle CAB\) and \(\angle DAC\) are not relevant for the given proportion.

Answer:

A. \(\angle ABC\cong\angle DBE\)