In Fig. 6.19, DE || AC and DF || AE. Prove that BF BE FE EC
In ∆ BCA, DE || AC
In Fig. 6. 19. DE || AC and DF || AE. Prove that BF/FE = BE/EC.
In Fig. 6. 19. DE is parallel to AC and DF is parallel to AE. Prove that BF/FE = BE/EC.
Solution
Solution:
We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.
In ΔABC
DE || AC
BD/AD = BE/EC .........(i)
In ΔABE
DF || AE
BD/AD = BF/FE ........(ii)
From (i) and (ii)
BD/AD = BE/EC = BF/FE
Thus, BE/EC = BF/FE
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