# 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 similarity criterion for two triangles.

In ΔAOB and ΔCOD

∠AOB = ∠COD (vertically opposite angles)

∠BAO = ∠DCO (alternate interior angles)

⇒ ΔAOB ∼ ΔCOD (AA criterion)

Hence, OA/OC = OB/OD

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

**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

**Summary:**

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

**☛ Related Questions:**

- In Fig. 6.36 QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
- S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
- In Figure 6.37, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
- In Fig. 6.38, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that: (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC.

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