D
C
A
B
M
R
P
Q
Proof :
ΔABC
~
ΔPQR
[Given]
AB
PQ
=
BC
QR
...(i) [corresponding sides
of similar triangles]
BC
=
2BD
…(ii) [D is the midpoint of seg BC]
QR
=
2QM
...(iii) [M is the midpoint of seg QR]
AB
PQ
=
2BD
2QM
[From (i), (ii), (iii)]
∴
∴
∠B
=
∠Q
...(v) [corresponding angles
of similar triangles]
AB
PQ
=
BD
QM
...(iv)
∴
If AD and PM are medians of triangles ABC and PQR respectively
where ΔABC ~ ΔPQR, prove that
AB
PQ
=
AD
PM
Hint :
To Prove :
ΔABD ~ ΔPQM
EX.6.3 (Q.16)
In ΔABD and ΔPQM,
∠B
=
∠Q
AB
PQ
=
BD
QM
∠B
=
∠Q
[From (iv)]
[From (v)]
D
C
A
B
M
R
P
Q
AB
PQ
=
BD
QM
...(iv)
∴
ΔABD
~
ΔPQM
[by SAS similarity criterion]
AB
PQ
=
AD
PM
∴
[corresponding sides of similar triangles]
∴
...(v)
Hint :
To Prove :
ΔABD ~ ΔPQM
Q. If AD and PM are medians of triangles ABC and PQR respectively
where ΔABC ~ ΔPQR, prove that
AB
PQ
=
AD
PM
Proof :
EX.6.3 (Q.16)