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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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Solution

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=BFGFGFBCCD=BFGF1
BCCD+1=BFCE [from Eq.(ii)]
BC+CDCD=BFCEBDCD=BFCE

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