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
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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
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