Question

In the given figure, D, E and F are mid-points of sides BC, CA and AB of A ABC. BE cuts DF at P and CF cuts DE at Q. Show that
(i) PQ Il EF
(ii) PQ = 1/4 * BC.​

Answer 1

Step-by-step explanation:

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