(2) Question 2: AE = BF

Proof: connecting PE, BP and AP, if the circumferential angles of the same arc are equal, you can get ∠ peq = ∠ pfq, ∠ pbq = ∠ paq, and if p is the midpoint of AB arc, you can get AP = BP.

Get △APE all equal to △BPF, and get AE = BF.

Question 1: isosceles right triangle

Proof: From ∠AQB, whose diameter is at right angles to the circumferential angle, to ∠EQF, to ∠ O', to ∠EPF.

Connecting PE, BP and AP, PE = PF can be obtained from ∠PEQ =∠PFQ∠PBQ =∠PAQ, AP = BP can be obtained from P to the midpoint of AB arc, and △APE is all equal to △BPF.