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.

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