Any solutions here?

The triangles AED and AFB are similar; if we take E' on AD and D' on AB such as AE' = AE and AD' = AD, then D'E' || BF and, by an appropriate dilating they may coincide, therefore the problem can be simplified : Let ABC be a triangle, BPC and BQC isosceles triangles (such that PQ is the perpendicular bisector of BC, with P and Q opposite A w.r.t. BC) having angle PBC + angle QBC = angle BAC; then AP and AQ are simmetrical w.r.t. angle bisector of the angle BAC. Proof of the simplified problem: Let M be the middle of the arc BC not containing A in the circumcircle of the triangle ABC, and T a variable point on this arc. Then BT and CT intersect the perpendicular bisector of BC at P and Q, with the property of our problem. Obviously BM is the bisector of the angle PBQ and the circle (ABC) is the Appolonius circle of the triangle PQB, hence AM is the bisector of the angle PAQ. Important notes: 1) In the given problem, building two triangles on each BF and DE similar to ABC and ADC, i.e. DPE and DP'E, then BQF and BQ'F, then the lines APQ and AP'Q' are simmetrical w.r.t. the bisector of the angle BAD. 2) The solution was given to me by my son, too busy to post it himself. Best regards, sunken rock

anyone mind helping me explain clearly how APQ and AP'Q' being symmetrical can lead to collinearity?

Solution: Let $O$ be the circumcenter of $BEFC$ then since $\angle FBA=\angle ADE$,$\angle FAB=\angle EAD$, $\angle BFO=\angle FBO=90-\angle FDB=\angle ACD=\angle PDE=\angle PED$ then $BAF\sim DAE$ $DPE\sim BOF$ and they have the same orientation so $ABOF\sim ADPE$ and $\angle DAP=\angle BAO$ Similarly $\angle DAQ=\angle EAO$ Since $\angle EAO=\angle BAO$ then $\angle DAP=\angle DAQ$ so $A,P,Q$ are collinear.