In figure, AD is a median of triangle ABC and M is the mid_ point of AD. Also, BM produced meets AC at N. Prove that AN=1/3AC.
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please also show the diagram of your question
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Given that: AD is the median and M is the mid-point of AD in triangle ABC
To Prove: AN =(1/3)AC
Construction: draw DG parallel to EN
Proof:
Basic Proportionality Theorem (BPT): If a line is drawn parallel to one side of the triangle to intersect the other two sides at distinct points. Then the other two sides are divided in the same ratio.
In,
Since, D is the mid- point of BC, so, BD = CD
Thus,
AG = CG …… (1)
In,
Since, E is the mid-point of AD, so, AE = ED
Thus,
AN = NG ……. (2)
From equation (1) and (2), we have,
AN=AG=CG
AN=1/3×AC
To Prove: AN =(1/3)AC
Construction: draw DG parallel to EN
Proof:
Basic Proportionality Theorem (BPT): If a line is drawn parallel to one side of the triangle to intersect the other two sides at distinct points. Then the other two sides are divided in the same ratio.
In,
Since, D is the mid- point of BC, so, BD = CD
Thus,
AG = CG …… (1)
In,
Since, E is the mid-point of AD, so, AE = ED
Thus,
AN = NG ……. (2)
From equation (1) and (2), we have,
AN=AG=CG
AN=1/3×AC
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