line segment df intersect the side AC of a triangle ABC at the point E such that Ei s the mid point of AC and angle AEF=AngleAFE.prove that BD/CD=BF/CE.
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Then after here is the solution
prove that BD/CD = BF/CE
Draw CG || DF In ΔBDF
CG || DF
∴ BD/CD = BF/GF .............(1) BPT
In ΔAFE
∠AEF=∠AFE
⇒AF=AE
⇒AF=AE=CE..............(2)
In ΔACG
E is the mid point of AC
⇒ FG = AF
∴ From (1) & (2)
BD/CD = BF/CE
I have just answered same question
Then after here is the solution
prove that BD/CD = BF/CE
Draw CG || DF In ΔBDF
CG || DF
∴ BD/CD = BF/GF .............(1) BPT
In ΔAFE
∠AEF=∠AFE
⇒AF=AE
⇒AF=AE=CE..............(2)
In ΔACG
E is the mid point of AC
⇒ FG = AF
∴ From (1) & (2)
BD/CD = BF/CE
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