Question

in the given diagram, line segment bd bisects angle abc. segment bd is extended to e, where line segment ec is parallel to line segment ab. write a two - column proof to show that \\(\frac{ab}{ad}=\frac{bc}{dc}\\).

what is the missing step?

a. \\(ab = ad\\), property of similar triangles
b. \\(\angle 2\cong\angle 4\\), property of isosceles triangle
c. \\(\angle 2\cong\angle 3\\), property of similar triangles
d. \\(ec = bc\\), property of isosceles triangle

Explanation:

To determine the missing step (step 7), we analyze the context. We know \( \triangle ABD \sim \triangle CED \) (step 4) and \( \frac{AB}{AD}=\frac{BC}{DC} \) (step 8, via substitution). Step 7 should relate to a property that connects angles or sides for the substitution. Looking at the options:

• Option A: \( AB = AD \) – No info suggests this, and it doesn’t fit the similarity/substitution flow.
• Option B: \( \angle 2 \cong \angle 4 \) – \( \angle 4 \) is vertical to \( \angle 3 \), but \( \angle 2 \) and \( \angle 4 \) don’t have a direct isosceles triangle relation here.
• Option C: \( \angle 2 \cong \angle 3 \) – \( \angle 1 \cong \angle 2 \) (given), and if \( \angle 2 \cong \angle 3 \), maybe related to triangle properties, but not directly for the proportion. Wait, no—wait, step 6 is \( \angle 2 \cong \angle 5 \) (alternate interior? Wait, \( EC \parallel AB \), so \( \angle 1 \cong \angle 5 \) (alternate), but \( \angle 1 \cong \angle 2 \), so \( \angle 2 \cong \angle 5 \). Then step 7: We need a relation for the sides. Wait, maybe the missing step is about \( EC = BC \) (isosceles triangle, so \( \angle 5 \cong \angle 2 \), and if \( EC = BC \), then \( \triangle BCE \) is isosceles, so \( \angle 5 = \angle 2 \), but no—wait, the correct approach: Step 7 should be a statement that allows \( \frac{AB}{AD}=\frac{BC}{DC} \) via substitution. If \( EC = BC \) (option D), then \( BC = EC \), and from similar triangles \( \frac{AB}{AD}=\frac{EC}{DC} \), so substituting \( EC \) with \( BC \) (since \( EC = BC \)) gives \( \frac{AB}{AD}=\frac{BC}{DC} \). Wait, no—let’s re-examine. The diagram has \( EC \parallel AB \), so \( \triangle ABD \sim \triangle CED \) (AA similarity: \( \angle 3 \cong \angle 4 \) (vertical angles), \( \angle 1 \cong \angle 5 \) (alternate interior, since \( EC \parallel AB \)), and \( \angle 1 \cong \angle 2 \) (given), so \( \angle 2 \cong \angle 5 \). Then, if \( EC = BC \), \( \triangle BCE \) is isosceles with \( EC = BC \), so \( \angle 5 = \angle 2 \) (which we already have), but more importantly, for the proportion: from similarity, \( \frac{AB}{AD}=\frac{EC}{DC} \). If \( EC = BC \) (option D), then substitute \( EC \) with \( BC \), giving \( \frac{AB}{AD}=\frac{BC}{DC} \), which is step 8. So the missing step (step 7) is \( EC = BC \), property of isosceles triangle (since if \( EC = BC \), \( \triangle BCE \) is isosceles with base \( BE \)). Wait, but let’s check the options again. Option D is \( EC = BC \), property of isosceles triangle. So that would mean \( BC = EC \), so in \( \triangle CED \) and \( \triangle ABD \), the ratio \( \frac{AB}{AD}=\frac{EC}{DC} \), and since \( EC = BC \), we substitute to get \( \frac{AB}{AD}=\frac{BC}{DC} \). So the missing step (step 7) is \( EC = BC \), property of isosceles triangle. Wait, but the options: D is \( EC = BC \), property of isosceles triangle. So that’s the missing step, because then \( EC = BC \), so we can replace \( EC \) with \( BC \) in the similarity proportion \( \frac{AB}{AD}=\frac{EC}{DC} \) to get \( \frac{AB}{AD}=\frac{BC}{DC} \).

Answer:

D. \( EC = BC \), property of isosceles triangle