# In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ ABD ~ ∆ ECF

**Solution:**

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

This is referred as AA criterion for two triangles.

In ΔABD and ΔECF

∠ADB = ∠EFC = 90º [∵ AD ⊥ BC and EF ⊥ AC]

∠ABD = ∠ECF [∵ In ΔABC, AB = AC ⇒ ∠ABC = ∠ACB]

⇒ ΔABD ~ ΔECF (AA criterion)

**Video Solution:**

## In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ ABD ~ ∆ ECF

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.3 Question 11:

In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ ABD ~ ∆ ECF

In above figure E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC. Hence proved that ΔABD ~ ΔECF