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In △ ABC, D is a point on the edge of BC, and BD: DC = 3:2, e is a point on the edge of AC. be and ad intersect point O, and Bo: OE = 4:1, the value of Ce: EA is obtained
The answer to this question is 8:3
AF: FB = BD: DC = Ce: EA = 2:3, the area of triangle ABC is 1, which is formed by the intersection of AD, be and CF
What is the area of the triangle? Trouble will help, urgent!
Let AD and CF intersect with m, ad and be intersect with N, be intersect with CF with P (1) connected CN, let △ BDN area be 2x (later area) △ DCN = 3x, △ Cen = 2A, △ AEN = 3A, there are 2x + 3x + 2A = 2 / 5, and 3x + 2A + 3A = 3 / 5, x = 4 / 95.2x = 8 / 95. (2) connected BM, let △ AEM = 2Y, △ BEM = 3Y, △ BDM = 2B, △ CDM = 3b, there are 2
In triangle ABC, BD: DC = 1:2, AE: EC = 1:3, calculate ABO area: CDOE area
E. D over AC, BC, be, ad over o
The area of ABO is 1 / 6 triangle area, the area of CDOE is 7 / 12 triangle area, and their ratio is 2:7
As shown in the figure, △ ABC is an equilateral triangle, and points D, e and F are points on line AB, BC and Ca respectively. (1) if ad = be = CF, is △ def an equilateral triangle? (2) if △ DEF is an equilateral triangle, is ad = be = CF true? Try to prove your conclusion
(1) It is proved that ∵ ABC is an equilateral triangle, and ∵ ad = be = CF, ∵ DB = EC = FA, (2 points) ∵ ADF ≌ △ bed ≌ △ CFE, (3 points) ∵ DF = de = EF, that is, ∵ DEF is an equilateral triangle
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