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Triangles
EXERCISE 6.3
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
In Fig. 6.39, ABC and AMP are two right , right angled at B and M respectively. Prove that: (i) ABC ~ AMP (ii) CA BC PA MP
Step-by-Step Solution
Key Concept: Use the AA (Angle‑Angle) similarity criterion. Both triangles are right‑angled, and they share the acute angle at \(A\). Hence the triangles are similar, which immediately gives the proportion of corresponding sides.
1. Identify the right angles\
\[\angle ABC = 90^{\circ},\qquad \angle AMP = 90^{\circ}.\]
These are given in the figure.
2. Show that the acute angle at \(A\) is common\
The line \(AB\) and the line \(AP\) lie on the same straight line (or \(AB\) is an extension of \(AP\)). Therefore\
\[\angle CAB = \angle CAP.\]
This is the second pair of equal angles.
3. Apply AA similarity\
Since two angles of \(\Delta ABC\) are equal respectively to two angles of \(\Delta AMP\) (one right angle and the acute angle at \(A\)), the triangles are similar by the AA criterion. Hence\
\[\Delta ABC \sim \Delta AMP.\]
4. Correspondence of sides\
From the similarity, the side opposite the right angle in each triangle corresponds to the side opposite the right angle in the other triangle. Thus the correspondence is:
\[\begin{aligned}
CA &\leftrightarrow PA,\\
BC &\leftrightarrow MP,\\
AB &\leftrightarrow AM.
\end{aligned}\]
5. Write the required proportion\
Using the correspondence of the sides we obtain\
\[\frac{CA}{BC}=\frac{PA}{MP}.\]
This is the statement to be proved in part (ii).
Hence both (i) and (ii) are established.
\[\angle ABC = 90^{\circ},\qquad \angle AMP = 90^{\circ}.\]
These are given in the figure.
2. Show that the acute angle at \(A\) is common\
The line \(AB\) and the line \(AP\) lie on the same straight line (or \(AB\) is an extension of \(AP\)). Therefore\
\[\angle CAB = \angle CAP.\]
This is the second pair of equal angles.
3. Apply AA similarity\
Since two angles of \(\Delta ABC\) are equal respectively to two angles of \(\Delta AMP\) (one right angle and the acute angle at \(A\)), the triangles are similar by the AA criterion. Hence\
\[\Delta ABC \sim \Delta AMP.\]
4. Correspondence of sides\
From the similarity, the side opposite the right angle in each triangle corresponds to the side opposite the right angle in the other triangle. Thus the correspondence is:
\[\begin{aligned}
CA &\leftrightarrow PA,\\
BC &\leftrightarrow MP,\\
AB &\leftrightarrow AM.
\end{aligned}\]
5. Write the required proportion\
Using the correspondence of the sides we obtain\
\[\frac{CA}{BC}=\frac{PA}{MP}.\]
This is the statement to be proved in part (ii).
Hence both (i) and (ii) are established.
Correct Answer: (i) \(\Delta ABC \sim \Delta AMP\) by AA similarity (right angle and common acute angle at \(A\)).\
(ii) From the similarity, \(\dfrac{CA}{BC}=\dfrac{PA}{MP}\).
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