in triangleABC , AB=c,BC=a,CA=b. if D is a point on produced BC so that AD bisects angleA externally then BD:DC=
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Answer:
MATHS
In △ ABC, ray AD bisect ∠A and intersect BC in D. if BC= a, AC= b and AB=c, prove that
(i) BD=
b+c
ac
(ii) DC=
b+c
ab
ANSWER
In △ABC, AD bisects ∠A
∴
AB
AC
=
BD
DC
⇒
c
b
=
BD
DC
⇒
c
b
+1=
BD
DC
+1 [ Adding 1 to both sides ]
⇒
c
b+c
=
BD
DC+BD
⇒
c
b+c
=
BD
BC
[ DC+BD=BC ]
⇒
c
b+c
=
BD
a
∴ BD=
b+c
ac
[ Hence proved ]
Similarly since AD bisects ∠A.
∴
AC
AB
=
DC
BD
⇒
b
c
=
DC
BD
⇒
b
c
+1=
DC
BD
+1
⇒
b
c+b
=
DC
BD+DC
⇒
b
c+b
=
DC
BC
⇒
b
c+b
=
DC
a
∴ DC=
b+c
ab
[ Hence proved ]
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