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.
2 answers:
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.
You might be interested in
A factorization of is
.
- The maximum number of roots of a polynomial of degree
is
- For a polynomial with real coefficients, the roots can be real or complex.
- The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if
is a root, then
is also a root.
If the roots of the polynomial are
, then it can be factorized as
.
Here, we are to find a factorization of . Also, given that
and
are roots of the polynomial.
Since is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.
Hence, and
are also roots of the given polynomial.
Thus, all the four roots of the polynomial , are:
.
So, the polynomial can be factorized as follows:
Therefore, a factorization of is
.
To know more about factorization, refer: brainly.com/question/25829061
#SPJ9