Line segment DF intersects the side AC of a triangle ABC at the point E such that E is the midpont of CA and angle AEF=angle AFE. Prove that BD/CD=BF/CE
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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
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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