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# D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that

i) BDEF is a parallelogram.

ii) ar (DEF) = 1/4 ar (ABC)

iii) ar (BDEF) = 1/2 ar (ABC)

**Solution:**

The midpoint theorem states that the line joining the mid-points of two sides of a triangle is parallel to the third side and half of its length.

If one pair of the opposite side in a quadrilateral is parallel and equal to each other then it is a parallelogram. Diagonals separate the parallelogram into two triangles of equal areas.

Let's construct a diagram according to the given question.

i) In ΔABC, D and E are the mid-points of side BC and AC respectively.

Therefore, the line joining points D and E will be parallel to line AB and also half of it as per the midpoint theorem.

The mid-point of AB is Fand E is the mid-point of AC

∴ ED ║AB and ED = 1/2 AB

Since ED ║AB.

Therefore, ED ║FB and ED = FB [F is the midpoint of AB]

Since one pair of the opposite side in quadrilateral BDEF is parallel and equal to each other, Therefore, BDEF is a parallelogram.

Similarly, we can prove, AEDF and CEFD are also parallelograms.

ii) Now, in parallelogram BDEF,

The diagonal DF of the parallelogram BDEF divides it in two triangles BDF and DEF of equal areas.

∴ ar (ΔDEF) = ar (ΔBDF)………..(i)

Similarly, DCEF is also parallelogram.

∴ ar (ΔDEF) = ar (ΔDEC)……….(ii)

Also, AEDF is a parallelogram.

∴ ar (ΔDEF) = ar (ΔAFE)……….(iii)

From equation (i), (ii) and (iii),

ar (ΔDEF) = ar (ΔBDF) = ar (ΔDEC) = ar (ΔAFE)……….(iv)

Now,

ar (ΔABC) = ar (ΔDEF) + ar (ΔBDF) + ar (ΔDEC) + ar(ΔAFE)……….(v)

ar (ΔABC) = ar (ΔDEF) + ar (ΔDEF) + ar (ΔDEF) + ar (ΔDEF) [Using (iv) & (v)]

ar (ΔABC) = 4 × ar (ΔDEF)

ar (ΔDEF) = 1/4 ar (ΔABC)

iii) ar (║gm BDEF) = ar (ΔBDF) + ar (ΔDEF) = ar(ΔDEF) + ar (ΔDEF)

ar (║ gm BDEF) = 2 ar (ΔDEF)

ar (║ gm BDEF) = 2× 1/4 ar (ΔABC)

ar (║ gm BDEF) = 1/2 ar (ΔABC)

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 9

**Video Solution:**

## D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that i) BDEF is a parallelogram. ii) ar (DEF) = 1/4 ar (ABC) iii) ar (BDEF) = 1/2 ar (ABC)

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.3 Question 5

**Summary:**

D, E, and F are respectively the mid-points of the sides BC, CA, and AB of triangle ABC, then BDEF is a parallelogram, Area of (DEF) = 1/4 Area of (ABC), and Area of (BDEF) = 1/2 Area of (ABC).

**☛ Related Questions:**

- In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:i) ar (DOC) = ar (AOB)ii) ar (DCB) = ar (ACB)iii) DA || CB or ABCD is a parallelogram.[Hint: From D and B, draw perpendiculars to AC.]
- D and E are points on sides AB and AC respectively of ∆ ABC such that ar (DBC) = ar (EBC). Prove that DE || BC
- XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF)
- The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that ar (ABCD) = ar (PBQR).

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