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Mathematics midterm examination questions
Solution: (1)∵PE⊥AB, ∠ B = 60,

Therefore, in the right triangle PEB, BE= BP= BC=PC,

∴∠BPE=30,

∫∠EPF = 60,

∴FP⊥BC,

∠∠B =∠C = 60,BE=PC,∠PEB=∠FPC=90,

∴△BEP≌△FPC,

∴BE=PF,

∫∠EPF = 60,

△ EPF is an equilateral triangle.

(2) e is EH⊥BC in h,

According to (1), FP⊥BC, FC=BP= BC=4, BE=CP= BC=2,

In triangular FCP, PFC = 90-∠ C = 30,

∫∠PFE = 60,

∴∠GFC=90,

In the right triangle FGC, ∠ c = 60, CF=4,

∴GC=2CF=8,

∴GB=GC-BC=2,

In the right triangle BEP, ∠ EBP = 60, BP=4,

∴PE=2,BE=2,

∴EH=BE? PE÷BP=,

∴S△GBE= BG? EH =;

(3)∫CF = 2,AC=6,

∴CF= AC=PC,

△ CPF is an equilateral triangle,

∴∠FPC=60,

∴∠BPE= 180-60-60=60,

∫∠b = 60,

∴△EBP is an equilateral triangle,

∴∠BEP=∠PFC=60,

∴∠PEA=∠PFA,

∠∠A =∠EPF = 60,

∴ Quadrilateral EPFA is a parallelogram,

∴PE=AF=6-2=4.