In Fig. 6.33, CM and RN are respectively the medians of ABC and PQR. If ABC ~ PQR, prove that : (i) AMC ~ PNR (ii) CM AB RN PQ (iii) CMB ~ RNQ
Step-by-Step Solution
Key Concept: In similar triangles, the ratios of corresponding sides are equal. The midpoint of a side divides the side into two equal parts, therefore a median of a triangle is a line joining a vertex to the midpoint of the opposite side. Consequently, the median of one triangle corresponds to the median of the similar triangle, and the triangles formed by a vertex, the midpoint of the opposite side and the third vertex are also similar.
1. Given data
* \(\Delta ABC \sim \Delta PQR\). Hence
$$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=k \quad (k>0).$$
* \(M\) is the midpoint of \(AB\) ⇒ \(AM = MB = \dfrac{AB}{2}\).
* \(N\) is the midpoint of \(PQ\) ⇒ \(PN = NQ = \dfrac{PQ}{2}\).
* \(CM\) and \(RN\) are medians.
2. Proof of (i) \(\Delta AMC \sim \Delta PNR\)
* In \(\Delta AMC\) the sides are \(AM, AC, CM\).
* In \(\Delta PNR\) the sides are \(PN, PR, RN\).
* Using the midpoint relations:
$$\frac{AM}{PN}=\frac{\dfrac{AB}{2}}{\dfrac{PQ}{2}}=\frac{AB}{PQ}=k.$$
* From similarity of the original triangles:
$$\frac{AC}{PR}=k \quad\text{and}\quad \frac{CM}{RN}=k$$
(the last equality follows because a median is a line joining a vertex to the midpoint of the opposite side; corresponding medians are in the same ratio as the corresponding sides).
* Hence the three pairs of corresponding sides of \(\Delta AMC\) and \(\Delta PNR\) are in the same ratio \(k\). By the S.S.S. criterion, the two triangles are similar.
3. Proof of (ii) \(\dfrac{CM}{AB}=\dfrac{RN}{PQ}\)
* From step‑2 we already have \(\dfrac{CM}{RN}=\dfrac{AB}{PQ}\).
* Cross‑multiplying gives \(\dfrac{CM}{AB}=\dfrac{RN}{PQ}\).
* Thus the ratio of a median to its corresponding side is the same in both triangles.
4. Proof of (iii) \(\Delta CMB \sim \Delta RNQ\)
* The sides of \(\Delta CMB\) are \(CM, MB, CB\).
* The sides of \(\Delta RNQ\) are \(RN, NQ, RQ\).
* Using the midpoint relations and the similarity of the original triangles:
\[\frac{CM}{RN}=\frac{AB}{PQ}=k,\qquad \frac{MB}{NQ}=\frac{\dfrac{AB}{2}}{\dfrac{PQ}{2}}=\frac{AB}{PQ}=k,\qquad \frac{CB}{RQ}=\frac{BC}{QR}=k.\]
* All three pairs of corresponding sides are in the same ratio \(k\); therefore, by the S.S.S. criterion, \(\Delta CMB\) is similar to \(\Delta RNQ\).
5. Conclusion
* Hence, all three required statements are proved.
* The key steps rely on (i) the definition of a median, (ii) the property that corresponding medians of similar triangles are in the same ratio as the corresponding sides, and (iii) the S.S.S. similarity criterion.
Correct Answer: (i) \(\Delta AMC \sim \Delta PNR\)\; (ii) \(\dfrac{CM}{AB}=\dfrac{RN}{PQ}\)\; (iii) \(\Delta CMB \sim \Delta RNQ\).