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प्रश्न
AD is a median of side BC of ABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF: AC = 1 : 3.
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उत्तर
Construction: Draw DS ∥ BF, meeting AC at S.
Proof:
In ΔBCF, D is the mid-point of AC DS || BF.
∴ S is the mid-point of CF.
⇒ CS = SF ....(i)
In ΔADS, E is the mid-point of AD and EF || DS.
∴ F is the mid-point of AS.
⇒ AF = FS ....(ii)
From (i) and (ii), we get
AF = FS = SC
⇒ AC = AF = FS + SC
⇒ AC = AF + AF +AF
⇒ AC = 3AF
⇒ `"AF"/"AC" = (1)/(3)`
⇒ AF : AC = 1 : 3.
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