1, a parallel line passing through point P and BC, intersects the BQ extension line at point P'.

2. Cross the intersection point S of CP and BQ as the vertical intersection line of BC, the intersection line of BC at point T and the intersection line of PP' at point R. ..

Based on ∠ PCB = ∠ QBC, △BST and △CST are completely symmetrical.

That is, ST is the vertical center line of BC.

Because of PP'//BC, ST must also vertically divide the PP' line at point R.

That is △PRS? It is also completely symmetrical with △P'RS.

Among them, there is ∠ PP 'q = ∠ QBC; ?

Based on ∠QBC=∠BAC, ∠ PP 'q = ∠ BAC is obtained;

Really? ∠PP ' q =∠PAQ；

According to the theorem of equal chord and equal angle, the p' point must also be on the circumscribed circle of △APQ.

In other words, the PP' line is the chord of this circumscribed circle.

And the straight line RT is the perpendicular bisector of the straight line PP', then the center of the circumscribed circle of △APQ must be the perpendicular bisector of its chord PP', which is the RT line.

From the previous analysis, we can also see that the center of the circumscribed circle of △ABC must also be the vertical center line of its chord BC line, that is, RT line.

Therefore, RT line is also the connection line of two circumscribed circles.

Then there are the two intersecting lines of the circumscribed circle, the vertical midline BC.