In the figure, line segment DF intersects the side AC of a ΔABC at the point E such that E is the midpoint of CA and ∠AEF and ∠AFE. Prove that BDCD=BFCE.
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Given, ΔABC, E is the mid-point of CA and ∠AEF=∠AFE
To prove that BDCD=BFCE
Construction :Take a point G on AB such that CG II EF.
Since E is the midpoint of CA. ∴ CE = AE ………(i)
CG II EF and E is the midpoint of CA.
So, CE = GF ……(ii) [ by mid-point theorem]
Now, in ΔBCG and ΔBDF,
CG II EF ∴BCCD=BGGF [by basic proportionality theorem] ⇒BCCD=BF−GFGF⇒BCCD=BFGF−1 ⇒BCCD+1=BFCE [from Eq.(ii)] ⇒BC+CDCD=BFCE⇒BDCD=BFCE