In Fig. 9.20, D, E, F are the mid-points of BC, CA
and AB respectively. Prove that AD bisects EF
А
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Answer:
Since D and E are the mid-points of the sides BC and AB respectively of △ABC.
Therefore,
DE∣∣BA
⇒ DE∣∣FA........(i)
Since D and F are mid-points of the sides BC and AB respectively of △ABC.
∴ DF∣∣CA⇒DF∣∣AE.......(ii)
From (i), and (ii), we conclude that AFDE is a parallelogram.
Similarly, BDEF is a parallelogram.
Now, in △DEF and △ABC, we have
∠FDE=∠A [Opposite angles of parallelogram AFDE)
and, ∠DEF=∠B [Opposite angles of parallelogram BDEF]
So, by AA-similarity criterion, we have
△DEf∼△ABC
⇒
ARE(△ABC)
Area(△DEF)
=
AB
2
DE
2
=
AB
2
(1/2AB)
2
=
4
1
[∵DE=
2
1
AB]
Hence, Area(△DEF):Area(△ABC)=1:4
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