Read the proof. given: ab ∥ de prove: △acb ~ △dce 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. we can state ∠c ≅ ∠c using the reflexive property. therefore, △acb ~ △dce by the aa similarity theorem. sss similarity theorem. aas similarity theorem. asa similarity theorem.

(A) AA similarity

Step-by-step explanation:

Given: AB is parallel to DE.

To prove: △ACB is similar to △DCE.

Proof: It is given that AB is parallel to DE, thus because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal of the parallel lines.

Now, from △ACB and △DCE, we have

∠CED≅∠CBA (corresponding angles of transversal CB and are therefore congruent)

and ∠C ≅ ∠C (Reflexive property)

Thus, by AA similarity postulate,

△ACB is similar to △DCE

Hence proved.

Thus, option A is correct.

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (