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Q11:

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

Solution :

Given:

1. $\triangle ABC$ is an isosceles triangle with $AB = AC$.

2. $E$ is a point on the side $CB$ produced.

3. $AD \perp BC$ (which implies $\angle ADB = \angle ADC = 90^\circ$).

4. $EF \perp AC$ (which implies $\angle EFC = 90^\circ$).

To Prove:

$\triangle ABD \sim \triangle ECF$

A B C D E F

Step 1: Analyze the properties of $\triangle ABC$

Since $\triangle ABC$ is isosceles with $AB = AC$, the angles opposite to the equal sides are equal [Theorem: Angles opposite to equal sides of a triangle are equal].

Therefore, $\angle ABC = \angle ACB$ (or $\angle ABD = \angle ECF$).

Step 2: Identify the criteria for similarity

To prove $\triangle ABD \sim \triangle ECF$, we shall use the Angle-Angle (AA) similarity criterion, which states that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Step 3: Compare $\triangle ABD$ and $\triangle ECF$

In $\triangle ABD$ and $\triangle ECF$:

1. $\angle ABD = \angle ECF$ (Proved in Step 1, as $\angle ABC = \angle ACB$).

2. $\angle ADB = \angle EFC$ [Since $AD \perp BC$ and $EF \perp AC$, both are $90^\circ$].

Step 4: Conclusion of Similarity

Since two angles of $\triangle ABD$ are equal to two corresponding angles of $\triangle ECF$, by the AA similarity criterion:

$\triangle ABD \sim \triangle ECF$

Final Answer:

Hence, it is proved that $\triangle ABD \sim \triangle ECF$ by the AA similarity criterion.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


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