In a triangle ABC, AB = 3.9 cm, AC = 5.2 cm and bisector of angle A meets BC at D. If DC = 2.8 cm, then find BD.
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Answers
Given in △ABC, AD is the bisector of angle A
By internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides.
(i)
AC
AB
=
DC
BD
∴
4.2
5
=
DC
2.5
∴ DC=
5
2.5×4.2
∴ DC=2.1cm
(ii)
AC
AB
=
DC
BD
∴
AC
5
=
3
2
∴ AC=
2
5×3
∴ AC=7.5cm
(iii)
AC
AB
=
DC
BD
∴
4.2
3.5
=
2.8
BD
∴ BD=
4.2
3.5×2.8
∴ BD=2.33cm
(iv)
AC
AB
=
DC
BD
Let BD be x then DC becomes 6−x
∴
14
10
=
6−x
x
∴
7
5
=
6−x
x
∴ 30−5x=7x
∴ 12x=30
∴ x=2.5cm
∴ BD=2.5cm and CD=6−2.5=3.5cm
(v)
AC
AB
=
DC
BD
∴
4.2
AB
=
6
10−6
∴ AB=
6
4×4.2
∴ AB=2.8cm
(vi)
AC
AB
=
DC
BD
∴
6
5.6
=
3
BD
∴ BD=
6
5.6×3
∴ BD=2.8cm
⇒ BC=BD+CD=2.8+3=5.8cm
(vii)
AC
AB
=
DC
BD
∴
AC
5.6
=
6−3.2
3.2
[ AB=AD ]
∴ AC=
3.2
5.6×2.8
∴ AC=4.9cm
(viii)
AC
AB
=
DC
BD
Let BD be x then DC becomes 6−x
∴
6
10
=
12−x
x
∴
3
5
=
12−x
x
∴ 60−5x=3x
∴ 8x=60
∴ x=7.5cm
⇒ BD=7.5cm and CD=12−7.5=4.5cm