Domain and range of -3x^2+16x+3
1 answer:
You might be interested in
Answer:
ASA
Step-by-step explanation:
Given:
Two triangles ABC and EDC such that:
AB ⊥ BD and BD ⊥ DE
C is the midpoint of BD.
The two triangles are drawn below.
Since, AB ⊥ BD and BD ⊥ DE
Therefore, the two triangles are right angled triangle. The triangle ABC is right angled at vertex B. The triangle EDC is right angled at vertex D.
Since, point C is the midpoint of the line segment BD.
Therefore, C divides the line segment BD into two equal parts.
So, segment BC ≅ segment CD (Midpoint theorem)
Now, consider the triangles ABC and EDC.
Statements Reason
1. ∠ABC ≅ ∠CDE Right angles are congruent to each other
2. BC ≅ CD Midpoint theorem. C is midpoint of BD
3. ∠ACB ≅ ∠ECD Vertically opposite angles are congruent
Therefore, the two triangles are congruent by ASA postulate.
So, the second option is correct.
