In ∆ABC, CD is the median and E is the mid-point of CD and AE extended meets BC at F. AB, BC and CF are 5, 6 and 7 cm. Find CF.
Explanation:
Let us draw DG || AF as shown in the figure.
Since D is the mid-point of AB, AD = DB Also, since E is the mid-point of CD, CE = ED
In ∆CDG, since DG || EF, hence CF : FG = CE : ED = 1 : 1 ⇒ CF = FG ...(1)
In ∆BAF, since DG || AF, hence BG : GF = BD : DA = 1 : 1 ⇒ BG = GF ...(2)
From (1) and (2), we get BG = GF = FC
∴ BG = GF = FC = ⅓ BC = 2 cm.
Hence, option (c).
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