# The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium

**Solution:**

We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.

In quadrilateral ABCD as shown above,

Diagonals AC, BD intersect at ‘O’

Draw OE || AB

In ΔABC,

OE || AB

⇒ OA/OC = BE/CE (Basic Proportionality Theorem) ---------- (1)

But, OA/OB = OC/OD (given)

⇒ OA/OC = OB/OD ----------- (2)

From equations (1) and (2)

OB/OD = BE/CE

In ΔBCD,

OB/OD = BE/CE

OE || CD (Converse of Basic proportionality theorem)

We know that, OE || AB and OE || CD

Thus, AB || CD

Hence we can say ABCD is a trapezium as one pair of opposite sides AB and CD are parallel.

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 6

**Video Solution:**

## The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium

Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 10

**Summary:**

The diagonals of a quadrilateral ABCD intersect each other at the point such that AO/BO = CO/DO. Hence it is proved that ABCD is a trapezium.

**☛ Related Questions:**

- In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)
- E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
- In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
- In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.

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