Question 5: D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that

- BDFE is a parallelogram
- ar(DEF) = ¼ ar(ABC)
- ar(BDEF) = ½ ar(ABC)

**Answer:**

From mid-point theorem; BD||EF

BD = ½ BC (Because D is midpoint)

Hence, EF = BD

As EF = BD

So, BDFE is a parallelogram

Similarly, it can be proved that EFDC and AEDF are parallelograms.

As BD = CD = EF

Hence, ar(BDFE) = ar(EFDC)

In triangles BED and EFD;

BD = EF

DE = DE

So, from SSS theorem; ΔBDE ≈ ΔEFD

Similarly, it can be proven that ΔEFD ≈ ΔCDF

Similarly, it can be proven that ΔEFD ≈ ΔFEA

Thus, ΔBDE ≈ ΔEFD ≈ ΔCDF ≈ ΔFEA

Hence, ar(DEF) = ¼ ar(ABC)

As parallelogram BDFE is composed of two triangles

Hence, ar(BDFE) = ½ ar(ABC) proved.

Question 6: In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that

- ar(DOC) = ar(AOB)
- ar(DCB) = ar(ACB)
- DA||CB or ABCD is a parallelogram

(Hint: From D and B, draw perpendiculars to AC. )

**Answer:** (i) In triangles DOC and AOB;

DC = AB (given)

DO = BO (given)

Angle DOC = Angle AOB (Vertically opposite angles)

Hence, from SAS theorem; ΔDOC ≈ ΔAOB

Hence, ar(DOC) = ar(AOB)

(ii) In triangles DCB and ACB;

DC = AB (Given)

CB = CB (Common side)

Hence, from SSS theorem; ΔDCB ≈ ΔACB

Hence, ar(DCB) = ar(ACB) Proved

(iii) As opposite sides are equal hence, the quadrilateral ABCD is a parallelogram and DA||CB is proved.

Question 7: D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Prove that DE||BC.

**Answer:**

Since ar(DBC) = ar(EBC)

And these triangles have a common base, i.e. BC

Hence, they must be having the same altitude.

So, they are between same parallels.

Hence, DE||BC proved.

Question 8: 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)

**Answer:** BEYC is a parallelogram because EB||YC (Given EB||AC) and EY||BC (because XY ||BC)

In triangle AEB and parallelogram BEYC;

ar(AEB) = ½ ar(BEYC) (because triangle and parallelogram are between same parallels.)

Similarly, ar(ACF) = ½ ar(BXFC) (because triangle and parallelogram are between same parallels).

Now, ar(BEYC) = ar(BXFC) (because they are between same parallels)

Hence, ar(AEB) = ar(ACF) proved

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