what is the angle bisector Theorem proof it now with the help of well labelled figure
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Answered by
2
Heya User,
--> 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 ^_^
--> 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 ^_^
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Answered by
11
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.
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
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
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