Answer:
See Reasoning Below
Step-by-step explanation:
To prove that AB = BL, or in other words AB ≅ BL, let us consider the triangles CED and BEL. If we were to prove they were congruent, then by CPCTC ( corresponding parts of congruent triangle are congruent ) DC ≅ BL. As AB ≅ DC by " Properties of Parallelogram " it would be that through transitivity, AB ≅ BL / AB = BL;
[tex]m<CED = m<BEL,\\BE = CE,\\m< DCE = LBE\\\\\\CED = BEL,\\DC = BL,\\AB = BL - Proved[/tex]
Now for " part 2 " we can consider that AB = DC, from part 1. If AB = BL, then AL = 2 ( AB ) by the Partition Postulate. AB = DC, so we can also say that AL = 2 ( DC ) - Proved
See attachment in statement reasoning form for part 1;
