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

Question 16

Given: \(\triangle ABC \sim \triangle PQR\), and AD, PM are medians so that D is the midpoint of BC and M is the midpoint of QR.

To prove: \(\dfrac{AB}{PQ} = \dfrac{AD}{PM}\).

Step 1: Use similarity of the big triangles

Since \(\triangle ABC \sim \triangle PQR\), corresponding sides are proportional:

\[\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}. \tag{1}\]

Also, the correspondence of vertices is \(A \leftrightarrow P,\ B \leftrightarrow Q,\ C \leftrightarrow R\).

Step 2: Express the mid-segments using medians

Because AD is a median, D is the midpoint of BC:

\[BD = DC = \frac{BC}{2}.\]

Because PM is a median, M is the midpoint of QR:

\[QM = MR = \frac{QR}{2}.\]

From (1),

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

Hence,

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

Step 3: Show \(\triangle ABD \sim \triangle PQM\)

Now consider \(\triangle ABD\) and \(\triangle PQM\):

• From (1) and (2) we have

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

Also from (1), taking the first and third ratios,

\[\frac{AB}{PQ} = \frac{AC}{PR}.\]

But AD and PM join corresponding vertices A–C and P–R at their midpoints, so by similarity of the big triangles, the segment joining a vertex to the midpoint of the opposite side scales in the same ratio; this gives

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

More formally, since triangles \(ABC\) and \(PQR\) are similar with scale factor \(k = \dfrac{AB}{PQ}\), every corresponding length (including medians) is multiplied by the same factor. Thus,

\[AD = k \cdot PM,\]

so

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

Step 4: Conclude the required result

Therefore,

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

Hence proved.