# Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD

**Solution:**

Let's draw a trapezium ABCD with AB || DC.

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This is referred to as AA criterion for two triangles.

In ΔAOB and ΔCOD (vertically opposite angles)

∠AOB = ∠COD (alternate interior angles)

∠BAO = ∠DCO (alternate interior angles)

⇒ ΔAOB ∼ ΔCOD (AA criterion)

Hence, OA/OC = OB/OD

**Video Solution:**

## Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD

### Class 10 Maths NCERT Solutions - Chapter 6 Exercise 6.3 Question 3:

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD

For diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion it is proved OA/OC = OB/OD