Given: AE ≅ CE ; DE ≅ BE Prove: ABCD is a parallelogram. Parallelogram A B C D is shown. Diagonals are drawn from point A to point C and from point D to point B and intersect at point E. Line segments A E and E C are congruent. Line segments D E and E B are congruent. We have that AB || DC. By a similar argument used to prove that △AEB ≅ △CED, we can show that △ ≅ △CEB by. So, ∠CAD ≅ ∠ by CPCTC. Therefore, AD || BC by the converse of the theorem. Since both pair of opposite sides are parallel, quadrilateral ABCD is a parallelogram.

1.AED

2.SAS

3.ACB

4.Aternate interior angles

Step-by-step explanation:

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0