In the ΔABC, AD is the median through A and E is the mid-point of AD. BE produced meets AC at F. Prove that AF = 
AC.
Given: ΔABC in which E is the mid-point of the median AD.
To Prove: AF =
AC
Construction: Draw DG ïï BF, cutting AC at G.
Proof: In the ΔADG, EF ïï DG and E is the
mid-point of AD.
... F is the mid-point of AG (mid-point theorem)
... AF = FG...(i)
Now in the ΔBCF, DG ïï BF and D is the mid-point of BC.
... G is the mid-point of CF (mid-point theorem)
... FG = GC...(ii)
From Eqns (i) and (ii),
we get
AF = FG = GC
Since AC = AF + FG + GC
... AC = 3AF
... AF =AC
To Prove: AF =
ACConstruction: Draw DG ïï BF, cutting AC at G.
Proof: In the ΔADG, EF ïï DG and E is the
mid-point of AD.
... F is the mid-point of AG (mid-point theorem)
... AF = FG...(i)
Now in the ΔBCF, DG ïï BF and D is the mid-point of BC.
... G is the mid-point of CF (mid-point theorem)
... FG = GC...(ii)
From Eqns (i) and (ii),
we get
AF = FG = GC
Since AC = AF + FG + GC
... AC = 3AF
... AF =
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AC.
