In triangle ABC shown below, DE is parallel to AC. The following two-column proof with missing statements and reasons proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally: Which statement and reason accurately completes the proof? A. 3. ∠BDE ≅ ∠BAC; Corresponding Angles Postulate 4. ∠A ≅ ∠C; Isosceles Triangle Theorem B. 3. ∠BDE ≅ ∠BAC; Alternate Interior Angles Theorem 4. ∠A ≅ ∠C; Isosceles Triangle Theorem C. 3. ∠BDE ≅ ∠BAC; Corresponding Angles Postulate 4. ∠B ≅ ∠B; Reflexive Property of Equality D. 3. ∠BDE ≅ ∠BAC; Alternate Interior Angles Theorem 4. ∠B ≅ ∠B; Reflexive Property of Equality

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oeerivona oeerivona · Answer 1 · 2020-08-16T03:41:30+02:00

Answer:

The correct option is;

C. 3. ∠BDE ≅∠BAC, Corresponding Angles Postulate 4. ∠B ≅ ∠B Reflexive Property of Equality

Step-by-step explanation:

The two column proof can be written as follows;

Statement, Reason

1. [tex]\overline{DE} \parallel \overline{AC}[/tex], Given

2. [tex]\overline {AB}[/tex] is a transversal , Conclusion from statement 1.

We note that ∠BDE and ∠BAC are on the same side of the transversal relative to the parallel lines, and are therefore, corresponding angles.

Therefore, we have;

3. ∠BDE ≅∠BAC, Corresponding Angles Postulate

Also

4. ∠B ≅ ∠B, Reflexive Property of Equality

In the two triangles, ΔABC and ΔDBE, we have ∠BDE ≅∠BAC and ∠B ≅ ∠B,

From ∠BDE + ∠B + ∠BED = 180°

∠BAC + ∠B + ∠BCA = 180°

Therefore, ∠BED = ∠BCA Substitution property of equality

Which gives;

5, ΔABC ~ ΔDBE, Angle Angle Similarity Postulate

6. BD/BA = BE/BC, Converse of the Side-Side-Side Similarity Theorem