So ∠ BAC = ∠ BPC, ∠ ABC = ∠ CPA = 60.
So △ABC is an equilateral triangle.
2 because ∠BAC=∠BPC, ∠PQB=∠AQC.
So △PQB and △AQC are similar, and the similarity ratio is K.
So Pb = AC × K, BQ = QC× k.
In the same way; In a similar way
△APQ is similar to△△ bqc, and the similarity ratio is H.
So AP = BC × H, AQ = QC × H.
So AP: Pb = BC× H: (AC× K) = H: K.
AQ:QB=QC×H:(QC×K)=H:K
So AP:PB=AQ:QB
3 do PG vertical BC through point p.
Then PG=GC
Because the area of △ABC is 4√3.
So BC=4
Let PG be X.
According to trigonometric function
X:(4-X)=2+√3
X=(8+4√3):( 1+2+√3)
According to trigonometric function
Pc = ((8+4 √ 3): (1+2+√ 3)): tan 45 or (√2:2).