F4 math problems ( Circles )

Join PC and QB and they meet at Z. Suppose BQ and AC meet at D and PC and AB meet at G.

∠ PCB = ∠ PCA = x ( equal arcs, equal ∠s )

∠ ABQ = ∠ QBC = y ( equal arcs, equal ∠s )

∠ PCB = ∠ PQB = x ( ∠s in the same segment )

∠ QPC = ∠ QBC = y ( ∠s in the same segment )

So,

∠ PZB = ∠ OZC = x + y ( ext. ∠ of △ )

Considering △DZC,

∠ ZDA = 2x + y ( ext. ∠ of△ )

Considering △GZB,

∠ ZGA = x + 2y ( ext. ∠ of △ )

So by ext.∠ of △ again,

∠ DFQ = ∠ GEP = x + y

Then by vert. opp. ∠s,

∠ AEF = ∠ AFE = x + y

Hence AE = AF ( side opp. equal ∠s )

Therefore △AEF is an isosceles triangle.

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