(Q24) CM and RN are respectively the medians of similar triangles ∆ABC and ∆PQR. Prove that

i)∆AMC ∼ ∆PNR

ii)
CM
RN
 =
AB
PQ

iii) ∆CMB ∼ ∆RNQ

ABCMRPNQ

Solution :

Given : ∆ABC ∼ ∆PQR

CM is a median through C of ∆ABC.

RN is a median through R of ∆PQR.

i) ∆AMC ∼ ∆PNR

Proof : In ∆AMC and ∆PNR

PR
 =
AM
 and ∠A = ∠ [∵ In ∆ABC , ∆PQR
AC
PR
 =
AB
PO
 =
(1/2)AB
(1/2)PQ
 and M, N are the mid-points of AB and PQ ]

∴ ∆AMC ~ ∆PNR

[∵ SAS similarity condition]

ii)
CM
RN
 =
AB
PQ

Proof: From (i) we have

∆AMC ∼ ∆PNR

Hence,
PQ
 =
AM
 =
RN

[∵ Ratio of corresponding sides of two similar triangles are equal]

Thus,
CM
RN
 =
AM × 2
PN × 2

[Multiplying both numerator and the denominator by 2]

CM
RN
 =
AB
PQ
   [2AM = AB; 2PN = PQ]

iii) ∆CMB ∼ ∆RNQ

Proof : In ∆CMB and ∆RNQ

∠B = ∠ [Corresponding angles of △ABC and △PQR]

Also,
RQ
 =
BM
[ ∵
BC
RQ
 =
AB
PQ
 ⇒
BC
PQ
 =
(1/2)AB
(1/2)PQ
 ]

Thus, ∆CMB ∼ ∆RNQ by S.A.S similarity condition.