Respuesta :
Given that DE||AC, by similar triangles, we have;
- D. <CAB congruent to <EDB, <ACB congruent to <DEB
How can the missing step be found?
A part of the question that appears missing is; side DE||AC
The completed two column proof is presented as follows;
Statement. Reason
1. DE||AC 1. Given
2. <EDB congruent to <CAB
<DEB congruent to <ACB
Which can be written as; <CAB congruent to <EDB, <ACB congruent to <DEB
2. If two parallel lines are cut by a common transversal, the corresponding angles are congruent
3. ∆ABC ~ ∆DBE 3. by AA criterion for similarity
4.
[tex] \mathbf{\frac{ AB }{ CB} } = \frac{ DB }{ EB} [/tex]
4. Corresponding sides of similar triangles are proportional
5. AB = AD + DB
CB = CE + EB 5. Segment addition postulate
6. [tex] \mathbf{\frac{ AD + DB }{ CE + EB} } = \frac{ DB }{ EB} [/tex]
6. Substitution property of equality
7. [tex] \frac{ AD + DB }{ DB} = \frac{ CE + EB }{ EB} [/tex]
7. Division property
8. [tex] \frac{ AD }{ DB} + 1 = \frac{ CE }{ EB} + 1 [/tex]
[tex] \frac{ AD }{ DB} = \frac{ CE }{ EB} [/tex]
8. Subtraction property
Therefore, the missing step in the proof is the option;
- D. <CAB congruent to <EDB, <ACB congruent to <DEB
Learn more about similar triangles here;
https://brainly.com/question/2272738
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