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Question 1188287: Given triangle ABC with D the midpoint of side AB and E on side BC, F on side AC and G is the intersection of side DE and side BF. BE:EC = 2:3 and DG:GE = 5:8. What is the ratio of BG:GF?
Answer by yurtman(42)   (Show Source):
You can put this solution on YOUR website! Certainly, let's find the ratio of BG:GF.
**1. Utilize Ceva's Theorem:**
* Ceva's Theorem states that for any triangle ABC, if lines AD, BE, and CF intersect at a single point (in this case, point G), then:
(AF/FC) * (BD/DA) * (CE/EB) = 1
**2. Apply the given ratios:**
* BD/DA = 1/1 (since D is the midpoint of AB)
* BE/EC = 2/3
**3. Calculate AF/FC:**
* Using Ceva's Theorem:
(AF/FC) * (1/1) * (3/2) = 1
AF/FC = 2/3
**4. Use Menelaus' Theorem:**
* Menelaus' Theorem states that for any transversal line (in this case, line DE) that intersects the sides of a triangle (triangle ABC) at points D, E, and F, then:
(AD/DB) * (BE/EC) * (CF/FA) = 1
**5. Apply the known ratios and solve for BG/GF:**
* (1/1) * (2/3) * (3/2) * (BG/GF) = 1
BG/GF = 1
**Therefore, the ratio of BG:GF is 1:1.**
Let me know if you have any other questions or problems to solve!
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