The medians BE and CD of ∆ABC are produced upto F and G respec-
tively such that BE = EF and CD = DG. Prove that G, A and F are
collinear.
Answers
Step-by-step explanation:
It is given that CD and GD are medians of △
′
sABC and EFG respectively.
∴ 2AD=AB and 2FH=FE......(i)
It is also given that △ABC∼△FEG
∴
FE
AB
=
FG
AC
=
EG
BC
and, ∠A=∠F,∠B=∠E,∠C=∠G...........(ii)
Now,
FE
AB
=
FG
AC
=
EG
BC
⇒
2FH
2AD
=
FG
AC
=
EG
BC
[Using (i)]
⇒
FH
AD
=
FG
AC
=
EG
BC
.........(iii)
(I) In △
′
sADC and FHG, we have
FH
AD
=
FG
AC
[From (iii)]
and , ∠A=∠F
So, by SAS criterion of similarity, we have
△ADC∼△FHG [Hence proved]
(ii) we have,
△ADC∼△FHG [Proved above]
⇒
HG
DC
=
FH
AD
⇒
GH
CD
=
2FH
2AD
⇒
GH
CD
=
FE
AB
[∵ AB=2AD and FE=2FH] [Hence proved]
(iii) We have,
FE
AB
=
FG
AC
=
EG
BC
[From (i)]
Also,
GH
CD
=
FE
AB
[As proved above]
∴
GH
CD
=
EG
BC
.......(iv)
Again,
FE
AB
=
FG
AC
=
EG
BC
⇒
2HE
2DB
=
EG
BC
[∵ D and H are mid-points of AB and FE respectively]
⇒
HE
DB
=
EG
BC
From (iv) and (v), we have
GH
CD
=
EG
BC
=
HE
DB
⇒
GH
CD
=
HE
DB
=
EG
BC
⇒ △CDB∼△GHE [By SSS criterion of similarity] [Hence proved]