(1) Proof:

∫△ABC is an equilateral△

∴∠c=∠abe=60 ab = BC

∴∠BAE+∠AEB= 120

∠CBD+∠BDC= 120

BE = CD

∴△ABE is equal to △ △BCD(SAS).

∴∠BAE=∠CBD

AEB=∠BDC

∴∠DBC+∠AEB= 120

∴∠BPE=60

∴∠APO=60

(2) prove that:

∵ Quadrilateral ABCF is a square

∴AB=BC

∠ABC =∠C = 90°

∴∠BAE+∠AEB=90

∠CBD+∠BDC=90

BE = CD

∴△ABE is equal to △ △BCD(SAS).

∴∠BAE=∠CBD

AEB=∠BDC

∴∠CBD+∠AEB=90

∴∠BPE=90

∴∠APD=90

(3)

Prove:

The pentagon ABCMN is a regular polygon.

∴∠abe=∠bcd= 108 ab = BC.

∴∠BAE+∠AEB=72

∠CDB+∠ Central Business District =72

BE = CD

∴△ABE is equal to △ △BCD(SAS).

∴∠BAE=∠CBD

∠BEA=∠CDB

∴∠CBD+∠AEB=72

∴∠BPE= 108

∴∠APO= 108

2. regular polygon ABCF ... points on two adjacent sides with c as the vertex, BE=CD.

Then ∠APO= 180(n-2)/n

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