Q. CD & GH are respectively the bisectors of ∠ACB & ∠EGF
such that D & H lie on sides AB & FE of ΔABC & ΔEFG
respectively. ΔABC ~ ΔFEG.
Prove that :
ii) ΔDCB ~ ΔHGE
i)
CD
GH
=
AC
FG
Soln.
D
A
B
C
×
×
H
F
E
G
o
o
iii) ΔDCA ~ ΔHGF
ΔABC
~
ΔFEG
..... [given]
∠ACB
=
∠FGE
..... [corresponding angles
of similar triangles]
1
2
∴
∠ACB
=
∠FGE
..... (i)
..... (ii)
..... [CD bisects
∠ACB]
..... (iii)
..... [GH bisects
∠ACB]
1
2
CD & GH are respectively the bisectors of ∠ACB & ∠EGF
×
×
o
o
What can we say about
∠BCD and ∠ACD ??
They are equal
∠BCD = ∠ACD = ∠ACB
1
2
What can we say about
∠EGH and ∠FGH ??
They are equal
∠EGH = ∠FGH = ∠FGE
1
2
ΔABC ~ ΔFEG
To prove:
Whenever we write the
products in the form of a ratio,
triangle and
another triangle.
# Remember
Does CD & AC form one
triangle ??
Yes
Which triangle ??
Does GH & FG form one
triangle ??
Yes
Which triangle ??
Hint :
To prove:
Δ DCA
~
Δ HGF
i)
CD
GH
=
AC
FG
∠EGH = ∠FGH = ∠FGE
1
2
∠BCD = ∠ACD = ∠ACB
1
2
=
Multiplying both sides by
1
2
∠BCD
=
∠ACD
=
∠ACB
1
2
∠EGH
=
∠FGH
=
∠FGE
1
2
CD
GH
AC
FG
EX.6.3 (Q.10)
=
... (iv)
... from (i), (ii), (iii)
∠A
=
∠F
... (vi)
...[corresponding angles
of similar triangles]
In ΔADC & ΔFHG,
∠ACD
=
∠FGH
... from(iv)
∠A
=
∠F
... from (vi)
ΔDCA
~
ΔHGF
... [by AA similarity
criterion]
CD
GH
AC
FG
=
... [corresponding sides
of similar triangles]
In ΔDBC & ΔHEG,
=
... from (i), (ii), (iii)
∠B
=
∠E
... from (v)
ΔDCB
~
ΔHGE
... [by AA similarity criterion]
Hint :
To prove:
~
Δ DCA
Δ HGF
1
2
∠ACB
=
∠FGE
1
2
∠BCD
=
∠ACD
=
∠ACB
1
2
∠EGH
=
∠FGH
=
∠FGE
1
2
i
ii
iii
1
2
∠ACB
=
∠FGE
1
2
∠ACD
∠FGH
D
A
B
C
×
×
H
F
E
G
o
o
∠ACB
1
2
∠FGE
1
2
∠BCD
∠EGH
1
2
∠ACB
=
∠FGE
1
2
∠ACB
1
2
∠FGE
1
2
∠B
=
∠E
... (v)
EX.6.3 (Q.10)