NCERT Solutions
Class 10 - Mathematics - Chapter 6: TRIANGLES - Exercise 6.3

Question 11

Given: \(\triangle ABC\) is isosceles with \(AB = AC\). E lies on side CB. AD is perpendicular to BC and EF is perpendicular to AC.

To prove: \(\triangle ABD \sim \triangle ECF\).

Step 1: Mark the right angles

Since AD ⟂ BC, angle at D in \(\triangle ABD\) is a right angle:

\[\angle ADB = 90^\circ.\]

Since EF ⟂ AC, angle at F in \(\triangle ECF\) is a right angle:

\[\angle EFC = 90^\circ.\]

Thus,

\[\angle ADB = \angle EFC. \tag{1}\]

Step 2: Use the isosceles property \(AB = AC\)

In an isosceles triangle ABC with \(AB = AC\), the base angles at B and C are equal:

\[\angle ABC = \angle BCA. \tag{2}\]

Step 3: Relate angles in the two smaller triangles

In \(\triangle ABD\), angle at B is

\[\angle ABD = \angle ABC,\]

because BD lies along BC.

In \(\triangle ECF\), angle at C is

\[\angle ECF = \angle BCA,\]

because CE lies along CB and CF lies along CA.

Using (2):

\[\angle ABD = \angle ABC = \angle BCA = \angle ECF. \tag{3}\]

Step 4: Conclude similarity

From (1) and (3), we have two pairs of equal angles:

\[\angle ADB = \angle EFC, \quad \angle ABD = \angle ECF.\]

Therefore, by the AA similarity criterion,

\[\triangle ABD \sim \triangle ECF.\]

Hence proved.