【试题分析】来自台湾的一道初中数学竞赛题

文摘   2024-10-22 16:35   广东  
[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.




大佬说数学
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