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ABCD is a parallelogram and E is the midpoint of BC. The side DC is extended such that it meets AE, when extended, at F. Prove that DF=2DC?
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ABCD is a parallelogram and E is the midpoint of BC. The side DC is ex...
Proof: DF = 2DC
1. Given: ABCD is a parallelogram, E is the midpoint of BC, and DC is extended to meet AE at point F.
2. Proof:
3. Since E is the midpoint of BC, we can say that BE = EC.
4. In triangle ABE, by the Midpoint Theorem, we know that AF is parallel to BC and that F is the midpoint of AD.
5. Therefore, by the Midpoint Theorem again, we have DF = 2FC.
6. Now, in triangle ADF, let's consider the line segment DC. Since F is the midpoint of AD, we can say that DF = 2DC.
7. Hence, we have proven that DF = 2DC in the parallelogram ABCD.
8. Conclusion: DF is indeed equal to 2 times DC in the given parallelogram ABCD.
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ABCD is a parallelogram and E is the midpoint of BC. The side DC is extended such that it meets AE, when extended, at F. Prove that DF=2DC?
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