There are four residential buildings A, B, C and D in a residential area. After measurement, it is found that the distance between the three residential buildings A, C and D is equal, and ∠ ACB = 90.

There are four residential buildings A, B, C and D in a residential area. After measurement, it is found that the distance between the three residential buildings A, C and D is equal, and ∠ ACB = 90. The property plan is located in, Answer: (1) proof: from the meaning of the question: AC=CD=DA,

∫∠ACB = 90 degrees,

△ ACD is an equilateral triangle,

∵DE⊥AC，

∴DE vertically divides AC,

∴AE=CE，

∫∠ACB = 90 degrees,

∴∠ACE+∠BCE=∠CAE+∠B=90，

∴∠BCE=∠B，

∴be=ce；

∴ae=ce=be；

(2) Solution: ∫DE bisects AC vertically,

∴PC=PA，

∴pb+pc=pb+pa；

∴PB+PC is the smallest, that is, PB+PA is the smallest, that is, P, B and A are the smallest on the same straight line.

That is, the minimum when p is at e,

When point P is at point E, PB+PC=AB=35 meters.