In \(\triangle ABC \sim \triangle PQR\), CM and RN are medians. Prove \(\triangle AMC \sim \triangle PNR\) and \(\triangle CMB \sim \triangle RNQ\).
Show Hint
If two triangles are similar, the ratio of any two corresponding segments (medians, altitudes, bisectors) is equal to the ratio of their corresponding sides.
Step 1: Solving OR (B):
1. Given \(\triangle ABC \sim \triangle PQR\), so \(AB/PQ = BC/QR = AC/PR\) and \(\angle A = \angle P, \angle B = \angle Q\).
2. Since M and N are medians, \(AB = 2AM\) and \(PQ = 2PN\).
3. Thus, \(2AM/2PN = AC/PR \implies AM/PN = AC/PR\).
4. In \(\triangle AMC\) and \(\triangle PNR\), \(AM/PN = AC/PR\) and \(\angle A = \angle P\). By SAS similarity, \(\triangle AMC \sim \triangle PNR\).
5. Similarly, \(MB/NQ = BC/QR\) and \(\angle B = \angle Q\), so \(\triangle CMB \sim \triangle RNQ\) by SAS. Step 2: Final Answer (OR):
Similarities are proved using the ratio of medians being equal to the ratio of sides.
