As shown in the figure, D and E are two points on the edge BC of △ ABC, and BD = EC. Prove: ab + AC = AD + AE

I don't know what the graph of your triangle looks like, but if you use equilateral triangle to consider this problem, it is not provable, because the edge AB will be greater than the edge ad, and the angle ADB is obtuse. In a triangle, the edge corresponding to obtuse angle is greater than the edge corresponding to acute angle. You can see if you give less conditions, or you can send it to the graph

As shown in the figure, ab = AC, ad = AE

It is proved that if AF ⊥ BC is f, ∵ AB = AC (known), ∵ BF = CF (three in one), ad = AE (known), ∵ DF = ef (three in one), ∵ bf-df = cf-ef, that is BD = Ce (the property of the equation)

In ABC, ad is the middle line, G is the middle point of AD, connect BG and extend AC to e, calculate the value of AE / AC
Make use of similar triangles
daan

1/3

In the figure, AE: EC = 1:2, CD: DB = 1:4, BF: FA = 1:3, and the area of △ ABC is s = 1, then the area of quadrilateral AFHG is______ ．

Connecting AF, CG ∵ BF: AF = 1:3 ∥ let the area of △ BFH = x, then the area of △ AFH = 3x. Similarly, let the area of △ ahe = y, then the area of △ CEH = 2Y. From the meaning of the title, we can get: the area of △ Abe = 4x + y = 13, the area of △ ACF = 3Y + 3x = 34, solve the binary linear equations 4x + y = 133Y + 3x = 34, then we can get: x = 136, that is, the area of △ BFH = 136, let the area of △ AEG = a, then the area of △ CEG = 2A, let the area of △ CDG = B, then the area of △ BDG = 136 Area = 4B can be obtained from the title: area of △ ACD = 3A + B = 15 area of △ BCE = 5B + 2A = 23 solve the binary linear equations 3A + B = 155B + 2A = 23 get: a = 139 area of quadrilateral AFHG = △ Abe area - △ BFH area - △ AEG area = 13-136-139 = 131468, so the answer is: 131468

As shown in the figure, D and E are two points on the edge BC of △ ABC, and BD = EC. Prove: ab + AC = AD + AE