3.75 cm
In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD.
If AB = 5 cm, AD = 4 cm and AC = 3 cm.
Find (i) BC, (ii) DC, (iii) A(ΔACD) : A(ΔBCA)
Proof :
In ΔABC and ΔDAC,
∴
A
B
C
D
[Given]
[Common angle]
…(i) [By AA criterion]
AB
DA
=
BC
AC
∠ABC = ∠DAC
∠ACB
=
∠DCA
ΔABC
~
ΔDAC
5 cm
3 cm
4 cm
∴
5
=
AC
DC
4
=
BC
3
=
3
DC
…(ii)
∴
5
4
=
BC
3
[From (ii)]
∴
5
× 3
=
BC
× 4
∴
BC
=
15
4
∴
BC
=
3.75 cm
∴
In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD.
If AB = 5 cm, AD = 4 cm and AC = 3 cm.
Find (i) BC, (ii) DC, (iii) A(ΔACD) : A(ΔBCA)
Proof :
3.75 cm
A
B
C
D
5 cm
3 cm
4 cm
5
4
=
BC
3
=
3
DC
…(ii)
∴
5
4
=
3
DC
[From (ii)]
∴
5
× DC
=
3
× 4
∴
DC
=
12
5
∴
DC
=
2.4 cm
(iii) ΔDAC ~ ΔABC
ar(ΔDAC)
ar(ΔABC)
=
AD2
AB2
=
(4)2
(5)2
A(ΔACD)
A(ΔBCA)
=
16
25
∴
A(ΔACD) : A(ΔBCA)
=
16 : 25
∴
∴
A(ΔACD)
A(ΔBCA)
∴
ΔABC ~ ΔDAC …(i)
[From (i)]