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

Question 9

Given: \(\triangle ABC\) and \(\triangle AMP\) are right-angled at B and M respectively.

To prove: (i) \(\triangle ABC \sim \triangle AMP\); (ii) \(\dfrac{CA}{PA} = \dfrac{BC}{MP}\).

(i) Prove \(\triangle ABC \sim \triangle AMP\)

In \(\triangle ABC\),

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

In \(\triangle AMP\),

\[\angle AMR = 90^\circ, \text{ i.e. } \angle AM P = 90^\circ.\]

So one angle of each triangle is a right angle.

Also, both triangles share angle at A:

\[\angle BAC = \angle PAM \quad (\text{common angle at } A).\]

Thus we have two pairs of equal angles:

\[\angle ABC = \angle AMP = 90^\circ, \quad \angle BAC = \angle PAM.\]

Therefore, by AA similarity,

\[\triangle ABC \sim \triangle AMP.\]

(ii) Prove \(\dfrac{CA}{PA} = \dfrac{BC}{MP}\)

From the similarity \(\triangle ABC \sim \triangle AMP\), corresponding vertices are:

\[A \leftrightarrow A, \quad B \leftrightarrow M, \quad C \leftrightarrow P.\]

So the corresponding sides are:

• \(AC \leftrightarrow AP\) (hypotenuses)
• \(BC \leftrightarrow MP\) (sides opposite the equal acute angles at A)

Hence, the ratio of corresponding sides is equal:

\[\frac{CA}{PA} = \frac{BC}{MP}.\]

Thus: (i) \(\triangle ABC \sim \triangle AMP\) and (ii) \(\dfrac{CA}{PA} = \dfrac{BC}{MP}\) are proved.