As shown in the figure, it is known that a is a point on the extension line of the side RQ of the equilateral △PQR, and the angle APB = 120. The following conclusion is correct.

( 1)△PAQ ~△BPR；

(2)QR？ =AQ RB .

Prove:

1)∫△PQR is an equilateral△

∴∠QRP=∠PRQ=∠PQR=60

∫∠APB = 120

∴∠apq+∠bpr= 120-∠prq =∠pqr =∠apq+∠a = 60

∴∠BPR=∠A

∴∠b=∠prq-∠bpr=60-∠a =∠apq∠pqa =∠BRP = 120

∴△PAQ~△BPR。

2)∫△PAQ ~△BPR；

∴AQ/PR=PQ/RB

PR = PQ = QR

∴AQ/QR=QR/RB

∴QR？ =AQ RB .