Question 16 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.
Proof Since E is the midpoint of CA. ∴ CE = AE ………(i) In ΔACG, 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