Help me now please ive been in this question for a really long time
1 answer:
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Answer: AAA similarity.
Step-by-step explanation: CB is the transversal for the parallel lines AB and DE, and so by transverse property, we have ∠CED ≅ ∠CBA. Similarly, CA acts as a tranversal for the same pair of parallel lines AB and DE and using the same property, we can have ∠CDE ≅ ∠CAB. Now, in triangles CED and ABC, we have
∠CED ≅ ∠CBA,
∠CDE ≅ ∠CAB
and
∠DCE ≅ ∠ACB [same angle]
Hence, by AAA (angle-angle-angle) similarity,
△CED ~ △ABC.
Thus, the correct option is AAA similarity.
Answer:
Coordinates of Vertices of triangle ABC are A (-4,3) , B(4,4) and C(1,1).
As, DO is Dilation of Δ ABC by Scale factor of .
Vertices of A' B'C' are
So, Image Δ A'B'C' will be smaller than the Pre image Δ ABC.
The two triangles will be congruent.
AO is Dilated by a factor of half , so A'O' will be half of AO.
So, correct Statements are
1. AB is parallel to A'B'.
2.DO,1/2(x, y) =
The distance from A' to the origin is half the distance from A to the origin.
