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

Question 12

Given: In \(\triangle ABC\), AD is a median, so D is the midpoint of BC. In \(\triangle PQR\), PM is a median, so M is the midpoint of QR. Also, the three lengths are proportional:

\[\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}.\tag{1}\]

To prove: \(\triangle ABC \sim \triangle PQR\).

Step 1: Express the smaller sides BD and QM

Since AD and PM are medians,

\[BD = DC = \frac{BC}{2}, \quad QM = MR = \frac{QR}{2}.\]

Using the second ratio in (1):

\[\frac{BC}{QR} = \frac{AB}{PQ}.\tag{2}\]

So,

\[\frac{BD}{QM} = \frac{\dfrac{BC}{2}}{\dfrac{QR}{2}} = \frac{BC}{QR} = \frac{AB}{PQ}.\tag{3}\]

Step 2: Prove \(\triangle ABD \sim \triangle PQM\) (SSS)

From (1) and (3), we have

\[\frac{AB}{PQ} = \frac{AD}{PM} = \frac{BD}{QM}.\]

Thus the three pairs of corresponding sides of \(\triangle ABD\) and \(\triangle PQM\) are proportional. Hence, by the SSS similarity criterion,

\[\triangle ABD \sim \triangle PQM.\]

This gives equality of corresponding angles:

\[\angle BAD = \angle QPM, \quad \angle ABD = \angle PQM.\tag{4}\]

Step 3: Transfer these angles to the big triangles

Since D lies on BC and M lies on QR, we have

\[\angle ABC = \angle ABD, \quad \angle PQR = \angle PQM,\]

and from (4),

\[\angle ABC = \angle ABD = \angle PQM = \angle PQR.\tag{5}\]

Similarly, at the vertices A and P,

\[\angle BAC = \angle BAD, \quad \angle QPR = \angle QPM,\]

so from (4),

\[\angle BAC = \angle BAD = \angle QPM = \angle QPR.\tag{6}\]

Step 4: Conclude similarity of \(\triangle ABC\) and \(\triangle PQR\)

From (5) and (6), we have two pairs of equal angles:

\[\angle ABC = \angle PQR, \quad \angle BAC = \angle QPR.\]

Therefore, by the AA similarity criterion,

\[\triangle ABC \sim \triangle PQR.\]

Hence proved.