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# 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 similarity criterion for two triangles.

In ΔABD and ΔECF

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

∠ABD = ∠ECF [∵ In ΔABC, AB = AC which signifies ∠ABC = ∠ACB as angles opposite to equal sides are equal]

Thus we have ΔABD ~ ΔECF (AA criterion)

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

**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

**Summary:**

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

**☛ Related Questions:**

- Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Fig. 6.41). Show that ∆ ABC ~ ∆ PQR.
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- Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆ PQR.
- A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower

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