I suggest one another proof for "Triangles With Equal Area":
https://www.cut-the-knot.org/Curriculum/Geometry/EqualAreasIMO2007.shtml

Please see attach file.

Suppose AC<=AB then Q near C than P

Ra, Rb = orthogonal projections of R on CA, CB respectively
Qa, Qb = orthogonal projections of Q on RRa, RRb respectively
Pa, Pb = orthogonal projections of P on RRa, RRb respectively
O' = orthogonal projections of O on CR

POQ is isosceles triangle because angle(OQP) = angle(OPQ) = 90 - angle(ACB)/2
so PO' = QO', moreover O' is midpoint of chord CR, therefore QC = PR. It makes two right triangles CLQ = PPaR and QL = RPa. As constructed QL//RPa therefore QLPaR is parallelogram. Similarly PKQbR is also parallelogram.

Denote (ABC) = area of ABC
(QLR) = (QPaR) = (QQaR) - (QQaPa)
(PKR) = (PQbR) = (QQbR) - (QPQb) = (QQbR) - (QQbPb)
By symmetry of angle bisector CR, these triangles are conguent and we are done.

Best regards,
Bui Quang Tuan

Please note that in the proof we use important fact that P, Q are reflections each other in midpoint of CR. So the fact can be generalized as:
P, Q are any two points on CR and they are reflections each other in midpoint of CR.
The area of triangle bonded by R, this point and its orthogonal projection on one side AC or BC is equal area of triangle bonded by R, other point and its orthogonal projection on other side.
The proof is the same.

Best regards,
Bui Quang Tuan