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Q7(iii):

In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iii) $\triangle AEP \sim \triangle ADB$

Solution :

Given:

In $\triangle ABC$, $AD$ is the altitude to side $BC$ (i.e., $AD \perp BC$) and $CE$ is the altitude to side $AB$ (i.e., $CE \perp AB$). The altitudes $AD$ and $CE$ intersect at point $P$.

To Prove:

$\triangle AEP \sim \triangle ADB$

B C A D E P

Step 1: Identify the triangles to be compared.

We are considering $\triangle AEP$ and $\triangle ADB$.

Step 2: List the corresponding angles and properties.

In $\triangle AEP$ and $\triangle ADB$:

1. $\angle AEP = \angle ADB$

Justification: Since $CE \perp AB$, $\angle AEP = 90^\circ$. Since $AD \perp BC$, $\angle ADB = 90^\circ$. Thus, both angles are equal to $90^\circ$.

2. $\angle PAE = \angle DAB$

Justification: Both angles refer to the same angle $\angle A$ of the original triangle $\triangle ABC$. This is the common angle shared by both triangles.

Step 3: Apply the Similarity Criterion.

According to the AA (Angle-Angle) Similarity Criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Since we have established:

$\angle AEP = \angle ADB = 90^\circ$

$\angle PAE = \angle DAB$ (Common angle)

Therefore, by the AA similarity criterion:

$\triangle AEP \sim \triangle ADB$

Conclusion:

We have successfully demonstrated that the two triangles satisfy the conditions for similarity based on the equality of their corresponding angles.

Final Answer: Since $\angle AEP = \angle ADB = 90^\circ$ and $\angle PAE = \angle DAB$ (common angle), by AA similarity criterion, $\triangle AEP \sim \triangle ADB$.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


CBSE Solutions for Class 10 Mathematics Triangles


Chapters in CBSE - Class 10 Mathematics


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