NCERT Solutions
Class 10 - Mathematics - Chapter 6: TRIANGLES - Exercise 6.2

Question 10

Given: In quadrilateral ABCD, diagonals AC and BD intersect at O and

\[\frac{AO}{BO} = \frac{CO}{DO}.\]

To prove: \(ABCD\) is a trapezium, i.e. one pair of opposite sides is parallel (here we shall show \(AB \parallel CD\)).

Step 1: Consider triangles AOB and COD

Look at \(\triangle AOB\) and \(\triangle COD\):

• They share a pair of vertical angles: \(\angle AOB = \angle COD\) (vertically opposite angles).
• We are given the ratio of the sides around these angles:

\[\frac{AO}{BO} = \frac{CO}{DO}.\]

So we have: an equality of the ratios of the two sides including the angle and the included angle is equal.

Therefore, by the SAS similarity criterion,

\[\triangle AOB \sim \triangle COD.\]

Step 2: Use similarity to relate angles on AB and CD

From the similarity \(\triangle AOB \sim \triangle COD\), corresponding angles are equal. With the correspondence \(A \leftrightarrow C\), \(B \leftrightarrow D\), \(O \leftrightarrow O\), we get in particular:

\[\angle ABO = \angle CDO,\]

\[\angle BAO = \angle DCO.\]

Step 3: Conclude that AB is parallel to CD

Note that:

• \(\angle ABO\) and \(\angle CDO\) lie on lines AB and CD with the transversal BD.
• \(\angle BAO\) and \(\angle DCO\) lie on lines AB and CD with the transversal AC.

Since these pairs of corresponding angles are equal, they are alternate interior angles. Hence,

\[AB \parallel CD.\]

Conclusion: One pair of opposite sides of quadrilateral ABCD is parallel, so ABCD is a trapezium with \(AB \parallel CD\).