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

Show that

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)

**Answer**

**Answer**

In ΔABC,

AO is the median. (CD is bisected by AB at O)

∴ ar(AOC) = ar(AOD) --- (i)

also,

In ΔBCD,

BO is the median. (CD is bisected by AB at O)

∴ ar(BOC) = ar(BOD) --- (ii)

Adding (i) and (ii) we get,

ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)

⇒ ar(ABC) = ar(ABD)

In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.

If AB = CD, then show that:

In the given figure, 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.]

**Answer**

Let us draw DN ⊥ AC and BM ⊥ AC.

(i) In ΔDON and ΔBOM,

∠ DNO = ∠ BMO (By construction)

∠ DON = ∠ BOM (Vertically opposite angles)

OD = OB (Given)

By AAS congruence rule,

ΔDON ≅ ΔBOM

∴ DN = BM ... (1)

We know that congruent triangles have equal areas.

∴ Area (ΔDON) = Area (ΔBOM) ... (2)

In ΔDNC and ΔBMA,

∠ DNC = ∠ BMA (By construction)

CD = AB (Given)

DN = BM [Using equation (1)]

∴ ΔDNC ≅ ΔBMA (RHS congruence rule)

⇒ Area (ΔDNC) = Area (ΔBMA) ... (3)

On adding equations (2) and (3), we obtain

Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)

Therefore, Area (ΔDOC) = Area (ΔAOB)

(ii) We obtained,

Area (ΔDOC) = Area (ΔAOB)

⇒ Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)

(Adding Area (ΔOCB) to both sides)

⇒ Area (ΔDCB) = Area (ΔACB)

(iii) We obtained,

Area (ΔDCB) = Area (ΔACB)

If two triangles have the same base and equal areas, then these will lie between the same parallels.

∴ DA || CB ... (4)

In ΔDOA and ΔBOC,

∠ DOA = ∠ BOC (Vertically opposite angles)

OD = OB (Given)

∠ ODA = ∠ OBC (Alternate opposite angles)

By ASA congruence rule,

ΔDOA ≅ ΔBOC

∴ DA = BC ... (5)

In quadrilateral ABCD, one pair of opposite sides is equal and parallel (AD = BC)

Therefore, ABCD is a parallelogram.

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 ΔBCE and ΔBCD are lying on a common base BC and also have equal areas, ΔBCE and ΔBCD will lie between the same parallel lines.

∴ 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 E respectively, show that ar (ABE) = ar (ACF)

Answer

Given,

XY || BC, BE || AC and CF || AB

To show,

ar(ΔABE) = ar(ΔAC)

Proof:

EY || BC (XY || BC) --- (i)

also,

BE∥ CY (BE || AC) --- (ii)

From (i) and (ii),

BEYC is a parallelogram. (Both the pairs of opposite sides are parallel.)

Similarly,

BXFC is a parallelogram.

Parallelograms on the same base BC and between the same parallels EF and BC.

⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) --- (iii)

Also,

△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.

⇒ ar(△AEB) = 1/2ar(BEYC) --- (iv)

Similarly,

△ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.

⇒ ar(△ ACF) = 1/2ar(BXFC) --- (v)

From (iii), (iv) and (v),

ar(△AEB) = ar(△ACF)