[Analysis and Solution] This is a question from a middle school math competition in Taiwan.This problem is relatively difficult. How can we find a way to solve it? Here are my thoughts:We can try exploring and analyzing from special cases to general cases to find a solution. Let's start by making triangle ABC, which is inscribed in a circle, an isosceles triangle with AB = AC, as shown in the diagram below: Clearly,△AKL≌△AML，△BKL≌△CML，△KLN≌MLN. To prove that the area of quadrilateral AKNM equals the area of △ABC, from the previous area relationships, it’s evident that we only need to show that the area of △KLN equals the area of △BKL.This means we need to show that KL // BN. Thus we draw line BN. Since BN ⊥ AB and KL ⊥ AB, it follows that KL // BN. In this special case (when △ABC is an isosceles triangle), we constructed a parallel line BN, transforming the area of △KLN into that of △BKL, thereby solving the problem.From this insight, we can obtain a solving strategy for the general case of triangle △ABC:From point N, we draw perpendiculars to lines AB and AC, with foot points labeled as E and F, respectively, and connect LE and LF, as shown in the diagram below:It is straightforward to prove that △LKE ≅ △LMF and that points E, L, and F are collinear.Thus, we have: [△KLN]=[△LKE] and [△LMN]=[△LMF] To prove our goal, it suffices to show that: [△BLE]=[△CLF]By connecting BN and FN, as shown in the diagram, we can use the HL theorem to determine that △BEN ≅ △CFN. From this, we conclude that: BE=CF Since ( LK = LM ), it follows that:[△BLE]=[△CLF].Thus, the problem is proven.
