文[1]先给出了一个定理:在△ABC中,AB,BC,CA三边上各任取一点分别为D,E,F,则nS△ADF+nS△BDE+nS△CEF≤3n2n-22S△ABC(图1),然后用它来解决一类三角形面积问题.如文[1]中例1:在△ABC中,AB,BC,CA上各有动点分别为D,E,F,当S△ABC=5,S△BDE=1,S△EFC=3,求△ADF面积的最大值与△DEF面积的最小值.以及例4:在△ABC中,AB,BC,CA上各有动点分别为D,E,F,当S△ADF=1,S△BDE=4,S△EFC=9,求 In [1], a theorem is given first: In △ ABC, AB, BC, CA on the three sides of each take a point for the D, E, F, then nS △ ADF + nS △ BDE + nS △ CEF ≤ 3n2n -22S △ ABC (Figure 1), and then use it to solve a triangular area problem. As in [1] in Example 1: In ABC, AB, BC, CA have moving point respectively D, E, F and the maximum value of ADF area and the minimum value of Δ DEF area when S Δ ABC = 5, S Δ BDE = 1, S Δ EFC = 3, and Example 4: In Δ ABC, AB, BC, CA On each of the moving points were D, E, F, when S △ ADF = 1, S △ BDE = 4, S △ EFC = 9, seeking