Question

State and prove angle bisector theorem ( external bisector).

Answer 1

Angle Bisector Theorem

Angle bisectors in a triangle have a characteristic property of dividing the opposite side in the ratio of the adjacent sides. More accurately,

Let AD - with D on BC - be the bisector of ∠A in ΔABC. If b = AC, c = AB, m = CD, and n = BD, then

b/c = m/n.

Jelena Nikolin from Serbia has graceously supplied several proofs. Below, (Δ) denotes the area of triangle Δ.

Proof 1

Here is a proof that does not appeal to the similarity of triangles. Instead, we'll argue that

(ACD) : (ABD) = m : n,

because the two triangles share the altitude from A.

On the other hand, point D is equidistant from the sides b and c (it belongs to the angle bisector), so altitudes of the smaller triangles from D are equals. These are denoted h:

(ACD) : (ABD) = bh/2 : ch/2 = b : c,

implying m : n = b : c.

Proof 2

Let B' and C' are foots of the perpendicular from B and C to the angle bisector. Triangles ABB' and ACC' are similar - the two are right and have equal acute angles at A. So

AB : AC = BB' : CC'. (*)

Further, triangles BB'D and CC'D are similar too so

BB' : CC' = BD : CD. (**)

From (*) and (**) we have AB : AC = BD : CD.

Remark: These two proofs are not original, but do have educational value.

Proof 3

Let E be the intersection of AD and the line parallel to AB through C. ∠AEC = ∠BAE (Transversal theorem: the line that cuts two parallels, cuts it under equal angles), meaning that ΔACE is isosceles and thus AC = CE. In addition, triangles ABD and ECD are similar, implying

AB : CE = BD : CD.

Together with CE = AC, we obtain the required AB : AC = BD : CD.

Proof 4

This proof is based on the following lemma that is of interest in its own right:

Let B' is the point on the segment AC where B'D||AB. Let C' is point on the segment AB such that C'D||AC. Then AB'DC' is a rhombus.Proof of Lemma∠DAC = ∠ADC' = ∠DAB = ∠ADB' = α/2 (transversal). Now, triangles ADB' and ADC' are congruent by ASA, soB'D = C'D = p. Since quadrilateral AB'DC' is a parallelogram with each sides equal, it is a rhombus.Using Thales' theorem, we have: B'D : AB = CD : BC orp : c = m : a (*)and C'D : AC = BD : BC, orp : b = n : a (**).Combining (*) and (**), we end up with m : n = b : c.

Answer 2

Here's the answer 

Refer the attachment for the figure

Theorem : 
The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Given:​ 
In ∆ ABC , side AD bisects Angle BAC such that B - D - C.

${\underline {\bf {To \: prove : }}}$

$\sf{ \frac{AB}{AC} = \frac{BD}{DC}}$ 

Construction :

Draw seg DE perpendicular to side AB and seg DN perpendicular to side AC.

Proof :

Point D lies on the bisector of Angle BAC
By angle bisector Theorem, DE = DF ... (1)

Now,
Ratio of areas of the two triangles is equal to the ratio of the products of their bases and corresponding heights.

$\sf{\frac{A(ADB)}{A(ADC)} = \frac{ (AB \times DE)}{ (AC \times DF )}}$

From 1, 

$\sf{\frac{A(ADB)}{A(ADC)} = \frac{ (AB )}{ (AC )}...(2)$

Also,∆ADB and ∆ADC have common vertex A 
and their bases BD and DC lie on the same line BC. So their heights are equal.

Area of triangles with equal heights are proportional to their corresponding bases.

${\sf{\frac{A( ∆ADB)}{A(∆ ADC)} = \frac{BD}{DC}}$

From 2 and 3 ,

We get 

${\sf{\frac{ AB }{AC} = \frac{BD}{DC}}$