Answer: AA similarity theorem.
Step-by-step explanation:
Given : AB ∥ DE
Prove: ΔACB ≈ ΔDCE
We are given AB ∥ DE. Because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal of the parallel lines. Angles CED and CBA are corresponding angles of transversal CB and are therefore congruent, so ∠CED ≅ ∠CBA.
Also ∠C ≅ ∠C using the reflexive property.
Therefore by AA similarity theorem , ΔACB ≈ ΔDCE
- AA similarity theorem says that if in two triangles the two pairs of corresponding angles are congruent then the triangles are similar .
x = 1
5(1)= -4y + 4
5= -4y + 4
-4. -4 1 = -4y
y = -¼
x = 2...
10 = -4y + 4 6 = -4y y = -3/2
x = 3..
15 = -4y + 4 11 = -4y y = -11/4
Answer:
l × b is the correct answer for this question
4(-8+4.5)
-32+18
-14
6.25+(-8.25)
-2
-14/-2
7
The answer is 7.
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