# 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.]

**Solution:**

We can draw a perpendicular from vertices B and D on diagonal AC which will help us to make congruent triangles and we know that congruent triangles are always equal in areas.

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

Let us construct DN ⊥ AC and BM ⊥ AC.

i) In ∆DON and ∆BOM,

∠DNO = ∠BMO = 90° (By construction)

∠DON = ∠BOM (Vertically opposite angles are equal)

OD = OB (Given)

By AAS congruence rule, ΔDON ≅ ΔBOM

DN = BM (By CPCT) ... (1)

We know that congruent triangles have equal areas.

Area (ΔDON) = Area (ΔBOM) ...(2)

Now, In ΔDNC and ΔBMA,

∠DNC = ∠BMA = 90° (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

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

Therefore, Area (ΔDOC) = Area (ΔAOB)

ii) We obtained,

Area (ΔDOC) = Area (ΔAOB)

Now, adding Area (ΔOCB) to both sides

∴ ar (ΔDOC) + ar (ΔOCB) = ar (ΔAOB) + ar (ΔOCB)

∴ Area (ΔDCB) = Area (ΔACB)

iii) We obtained,

Area (ΔDCB) = Area (ΔACB)

If two triangles are having the same base and also having equal areas, then these triangles must be lie between the same parallels.

∴ DA || CB ... (4)

In quadrilateral ABCD, one pair of opposite sides are equal (AB = CD) and the other pair of opposite sides are parallel (DA || CB).

Therefore, ABCD is a parallelogram.

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

**Video Solution:**

## 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 a parallelogram. [Hint: From D and B, draw perpendiculars to AC.]

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.3 Question 6

**Summary:**

If diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD, and AB = CD, then ar (DOC) = ar (AOB), ar (DCB) = ar (ACB), and DA || CB or ABCD is a parallelogram.

**☛ Related Questions:**

- 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).
- Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

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