Question

in triangle ABC, let AB=c,BC=a,CA=b.if D is a point on produced BC such that AD bisects angleA externally,then BD:DC=​

Answer 1

Step-by-step explanation:

Let, |DC|=x, so that |BD|=|BC|-|DC|=k-x.

According to the Angle Bisector Theorem, the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC.

|BD|/|DC|=|AB|/|AC|

(k-x)/x=m/l

x=l*k/(m+l)

Answer 2

Answer:

$\frac{BD}{DC} = \frac{BA}{AC}$

Concept:

In the ratio of the sides containing the angle, the external angle bisector of a triangle splits the opposing side externally.

Given

ΔABC

AB=c

BC=a

CA=b

AD is the external bisector

Find

The ratio of BD:DC

Solution

Draw CE ∥ DA meeting AB at E

Since, CE ∥ DA and AC is transversal, therefore,

∠ECA  =  ∠CAD (alternate angles)   .......(1)

and

∠CEA  =  ∠DAP (corresponding angles)     .......(2)

But AD is the bisector of ∠CAP,

∴ ∠CAD  =  ∠DAP    .......(3)

We know that, sides opposite to equal angles are equal.

Therefore,

∠CEA  =  ∠ECA

In ΔBDA, EC ∥ AD.

$\frac{BD}{DC} = \frac{BA}{AE}$

AE  =  AC,

$\frac{BD}{DC} = \frac{BA}{AC}$