It is possible, depending on what on earth AC and BC are!
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(a - 5b)(c - 5d)
You cannot prove it since it is not true for a general quadrilateral.
In this case you want to group the terms so they have at least two terms in common. First step group and rewrite it: abc + a'bc + a'b'c' + a'b'c + ab'c' + abc' = Use the rule Identities x(y+z)=xy+xz: bc(a+a') + a'b'(c'+c) + ac'(b'+b) = Use the rule Identities x+x'=1: bc (1) + a'b'(1) + ac'(1) = Use the rule Identities x(1) = x: bc+a'b'+ac'
6(b - ac + b2 - bc)
15 units