Question

In \(\triangle ABC \sim \triangle PQR\), CM and RN are medians. Prove \(\triangle AMC \sim \triangle PNR\) and \(\triangle CMB \sim \triangle RNQ\).

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.

Answer 1

Step 1: Given Information:
It is given that ΔABC ∼ ΔPQR.
Therefore, corresponding sides are proportional and corresponding angles are equal:
AB / PQ = BC / QR = AC / PR
∠A = ∠P
∠B = ∠Q

M and N are midpoints (since they are medians).
Hence,
AM = MB = AB/2
PN = NQ = PQ/2

Step 2: Proving ΔAMC ∼ ΔPNR:
Since AB / PQ = AC / PR
Replace AB = 2AM and PQ = 2PN:

2AM / 2PN = AC / PR
AM / PN = AC / PR

Now in triangles AMC and PNR:
AM / PN = AC / PR
∠A = ∠P (given from similarity of main triangles)

Thus, two sides are proportional and the included angle is equal.
By SAS similarity criterion:
ΔAMC ∼ ΔPNR

Step 3: Proving ΔCMB ∼ ΔRNQ:
From similarity of main triangles:
BC / QR = AB / PQ

Replace AB = 2MB and PQ = 2NQ:
2MB / 2NQ = BC / QR
MB / NQ = BC / QR

Now in triangles CMB and RNQ:
MB / NQ = BC / QR
∠B = ∠Q (given)

Again, two sides proportional and included angle equal.
By SAS similarity criterion:
ΔCMB ∼ ΔRNQ

Final Conclusion:
Using proportionality of medians derived from similar triangles and applying SAS similarity criterion, we prove:
ΔAMC ∼ ΔPNR
ΔCMB ∼ ΔRNQ
Hence, required similarities are proved.