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As shown in the figure, in ladder ABCD, ad ‖ BC, be bisects ∠ ABC, and be ⊥ CD is the first moving point on be. If BC = 6, CE = 2DE, then the maximum value of | pc-pa | is______ ．
Extend the extension line of Ba intersection CD to F, ∵ be bisecting ∵ ABC, ∵ FBE = ≌ CBE, ≁ CD, ∵ bef = ≁ BEC = 90 °, ∵ in △ FBE and △ CBE, ≌ be ≌ CBE (ASA), ≁ BF = BC = 6, EF = EC, ≁ be ⊥ CF, ≁ PC = PF (the distance from the point on the vertical bisector of the line segment to the two ends of the line segment is equal), that is | pc-pa | = | pf-pa | The shortest line segment between the two points is: | pf-pa | ≤ AF, that is, when the maximum value of | pc-pa | = AF, | EF = CE, CE = 2DE, | DF = de = 12ce = 14cf, ∥ ad ∥ BC, | AFD ∥ BFC, | afbf = fdcf = 14, | AF = 14bc = 14 × 6 = 32, that is, the maximum value of | pc-pa | is 32
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