Question

given \\( \triangle a b c \\) with \\( d e \\| a c \\) prove: \\( \frac{a d}{d b}=\frac{c e}{e b} \\) what is the missing step in this proof? a. \\( \angle c a b \cong \angle a c b, \angle e d b \cong \angle d e b \\) b. \\( \angle a d e \cong \angle d b e, \angle c e d \cong \angle e b d \\) c. \\( \angle c a d \cong \angle a c e, \angle a d e \cong \angle c e d \\) d. \\( \angle c a b \cong \angle e d b, \angle a c b \cong \angle d e b \\)

Explanation:

Since \(DE\parallel AC\), by the corresponding angles postulate, \(\angle CAB\) and \(\angle EDB\) are corresponding angles (so \(\angle CAB\cong\angle EDB\)), and \(\angle ACB\) and \(\angle DEB\) are corresponding angles (so \(\angle ACB\cong\angle DEB\)). This helps in establishing the similarity of \(\triangle ABC\) and \(\triangle DBE\) (by AA - Angle - Angle similarity criterion) which is a step towards proving \(\frac{AD}{DB}=\frac{CE}{EB}\) (using the property of similar triangles and ratio of sides).

Answer:

D. \(\angle CAB\cong\angle EDB,\angle ACB\cong\angle DEB\)