what is the angle bisector Theorem proof it now with the help of well labelled figure
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
--> Ummm, I'm sorry but I'm =_= soooo uncomfortable with Constructions.. So, I'll use a rather happy Geometry proof ^_^
--> Also, I use the notation [ ABC ] to define the area of ΔABC
--> Now, in ΔABC :->
--> [ ADB ] / [ ADC ] = BD / DC --> (i)
--> Note :-> [ Both triangles have a common height ]
Further, DE and DF are heights to AB and AC ..
--> Note :-> AD = AD || ∠EAD = ∠FAD || ∠AED = ∠AFD
=> ΔAED is congruent to ΔAFD --> { ASA Criteria }
=> DE = DF --> { c.p.c.t }
Further, [ ABD ] / [ ACD ] = { 1/2 * AB * DE } / { 1/2 * AC * DF }
==> [ ABD ] / [ ACD ] = AB / AC --> (ii)
Equating (i) and (ii) -->
---> AB / AC = BD / DC Proved ....
Well, since u've asked to prove the angle bisector theorem,
^_^ I presume that you know what it states :p
--> Plz let me know if uh wish to ask anything ^_^
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.
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.
From 1,
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.
From 2 and 3 ,
We get