Question

find the length of \\(\overline{ab}\\).
\\(\overline{ab} = \square\\)

Explanation:

Step1: Check similarity of triangles

Since \( \angle ABC=\angle AEC\) (corresponding angles) and \( \angle ACB=\angle ACE\) (common angle), \(\triangle ABC\sim\triangle AEC\) by AA (Angle - Angle) similarity criterion.

Step2: Use the property of similar triangles

For similar triangles \(\triangle ABC\) and \(\triangle AEC\), the ratios of corresponding sides are equal. That is \(\frac{AB}{AE}=\frac{BC}{EC}\). Also, since \(AE = AB + BE\) and \(BD\parallel AE\) (implied by the mid - segment - like appearance, but using the ratio of sides from similarity), \(\frac{BC}{BC + BE}=\frac{BD}{AE}\). But more straightforwardly, from \(\triangle ABC\sim\triangle AEC\), we know \(\frac{AB}{AB + BE}=\frac{BC}{BC + CD}\). Wait, another approach: Since \(BD\parallel AE\), by the basic proportionality theorem (Thales' theorem) \(\frac{BC}{AC}=\frac{BD}{AE}\). Also, \(AC=3 + 4=7\), \(EC = 12\). But using the ratio of sides of similar triangles \(\frac{AB}{AB + BE}=\frac{BC}{BC + CD}\) is wrong. Let's use the ratio of sides of \(\triangle ABC\) and \(\triangle AEC\) correctly.

Since \(BD\parallel AE\), \(\frac{BC}{AC}=\frac{BD}{AE}\). Wait, no, since \(\triangle ABC\sim\triangle AEC\) (because \(BD\parallel AE\) gives \(\angle CBD=\angle CAE\) and \(\angle CDB=\angle CEA\)), \(\frac{BC}{AC}=\frac{CD}{EC}\). Wait, no, the correct ratio for similar triangles \(\triangle CBD\) and \(\triangle CAE\) (since \(BD\parallel AE\)): \(\frac{CB}{CA}=\frac{CD}{CE}=\frac{BD}{AE}\). \(CB = 3\), \(CA=3 + x\) (where \(x\) is \(AB\)), \(CD = 4\), \(CE=12\). \(\frac{3}{3 + x}=\frac{4}{12}\)

Cross - multiply:

$$4(3 + x)=3\times12$$

$$12+4x = 36$$

$$4x=36 - 12$$

$$4x=24$$

$$x = 9$$

Answer:

\(9\)