NCERT Solutions
Class 10 - Mathematics - Chapter 6: TRIANGLES - Exercise 6.3

Question 3

Given: ABCD is a trapezium with \(AB \parallel CD\), and diagonals AC and BD intersect at O.

To prove: \(\dfrac{OA}{OC} = \dfrac{OB}{OD}\).

Step 1: Consider triangles \(\triangle AOB\) and \(\triangle COD\)

Look at triangles formed by the diagonals across the parallel sides:

• \(\triangle AOB\) has vertices on A, O, B.
• \(\triangle COD\) has vertices on C, O, D.

Angle 1: Since \(AB \parallel CD\) and BD is a transversal,

\[\angle ABO = \angle CDO \quad (\text{alternate interior angles})\]

Angle 2: At the intersection of diagonals,

\[\angle AOB = \angle COD \quad (\text{vertically opposite angles})\]

Thus, we have two pairs of equal angles:

\[\angle AOB = \angle COD, \quad \angle ABO = \angle CDO\]

So,

\[\triangle AOB \sim \triangle COD \quad (\text{AA similarity}).\]

Step 2: Use similarity to relate corresponding sides

From similarity \(\triangle AOB \sim \triangle COD\), the ratios of corresponding sides are equal. With the correspondence \(A \leftrightarrow C\), \(B \leftrightarrow D\), \(O \leftrightarrow O\), we get

\[\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}.\]

Hence, in particular,

\[\boxed{\dfrac{OA}{OC} = \dfrac{OB}{OD}}\]

Therefore, using triangle similarity, the required relation is proved.