Question

CM and RN are respectively the
medians of similar triangles ∆ABC and
∆PQR. Prove that
(1) ∆AMC ~ ∆PNR
(2)CM/RN = AB/PQ
(3) ∆CMB ~ ∆RNQ​

Answer 1

Answer:

ΔABC and ΔPQR

CM is the median of ΔABC and RN is the median of $ΔPQR

Also,

ΔABC∼ΔPQR

To Prove: ΔAMC∼ΔPNR

Proof:

CM is median of ΔABC

so, AN=MB=

2

1

AB......(1)

Similarly, RN is the median of ΔPQR

So, PN=QN=

2

1

PQ......(2)

Given,

ΔABC∼ΔPQR

PQ

AB

=

QR

BC

=

RP

CA

(Corresponding sides of similar triangle are proportional)

PQ

AB

=

RP

CA

2PN

2AM

=

RP

CA

{from (1) & (2)}

PN

AM

=

RP

CA

...........(3)

Also,

Since ΔABC∼ΔPQR

∠A=∠B (corresponding angles of similar triangles are equal)

In ΔAMC∼ΔPNR

∠A=∠P From (4)

PN

AM

=

RP

CA

from (3)

Hence by S.A.S similarly

ΔAMC∼ΔPNR

Hence proved... pls mark as brainlist