# In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area(ABC) / area(DBC) = AO/DO.

**Solution:**

In ΔABC, draw AM ⊥ BC

In ΔDBC, draw DN ⊥ BC

Now,

In ΔAOM and ΔDON

∠AMO = ∠DNO = 90 °

∠AOM = ∠DON (Vertically opposite angles)

⇒ ΔAOM ~ ΔDON (AA criterion)

⇒ AM / DN = OM / ON = AO / DO.............. (1)

Now,

Area of ΔABC = 1/2 × base × height = 1/2 × BC × AM

Area of ΔDBC = 1/2 × BC × DN

Area of ΔABC / Area of ΔDBC = (1/2 × BC × AM) / (1/2 × BC × DN)

Area of ΔABC / Area of ΔDBC = AM/DN

Thus, Area of ΔABC / Area of ΔDBC = AO/DO [From equation (1)]

**Video Solution:**

## In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area( ABC) / area(DBC) = AO/DO.

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.4 Question 3:

**Summary:**

In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, we have proved that area( ABC)/area(DBC) = AO/DO.