In ΔABC shown below, segment DE is parallel to segment AC: Triangles ABC and DBE where DE is parallel to AC The following two-co
lumn 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: Statement Reason 1. Line segment DE is parallel to line segment AC 1. Given 2. Line segment AB is a transversal that intersects two parallel lines. 2. Conclusion from Statement 1. 3. ∠BDE ≅ ∠BAC 3. Corresponding Angles Postulate 4. 4. 5. 5. 6. BD over BA equals BE over BC 6. Converse of the Side-Side-Side Similarity Theorem Which statement and reason accurately completes the proof? 4. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate 5. ∠B ≅ ∠B; Reflexive Property of Equality 4. ∠B ≅ ∠B; Reflexive Property of Equality 5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate 4. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate 5. ∠A ≅ ∠C; Isosceles Triangle Theorem 4. ∠A ≅ ∠C; Isosceles Triangle Theorem 5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate
2 answers:
You might be interested in
Idk if its right sorry
B is the answer
i simplified the theorem, but i hope this makes sense!
A box (cube) has all equal edges or sides (s), thus the volume (v) of the cube is:
So since each side or edge around is 3 in, then the length around the square bottom of the box = 4×3 = 12 in
So since each side or edge around is 3 in, then the length around the square bottom of the box = 4×3 = 12 in