How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of l

dumnealorjavgj 2022-04-06 Answered
How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length:
2 b c s ( s a ) b + c
I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.
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Answers (2)

reflam2kfnr
Answered 2022-04-07 Author has 17 answers
A method where no trigonometry is used.

Consider triangle A B C. Let A D, the angle bisector, intersect the circumcircle at L. Join L C. Consider triangle A B D and triangle A L C.

Triangle A B D is similar to triangle A L C (by A.A similarity theorem). Therefore,
A D A C = A B A L
i.e,
A D A L = A C A B
= A D ( A D + D L ) = A C A B
= A D A D + A D D L = A C A B  ... (1)
By power of point of point result:
A D D L = B D D C
B D = B C A B A B + A C
D C = B C A C A B + A C
In (1) ,
A D A D = A C A B B C 2 A B A C ( A B + A C ) 2
A D 2 = A C A B ( 1 B C 2 ( A B + A C ) 2 )
If A B = c , B C = a , A C = b:
A D 2 = b c ( 1 a 2 ( b + c ) 2 )

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Berghofaei0e
Answered 2022-04-08 Author has 3 answers
Assuming the usage of Trigonometry is allowed,

Let A D be the bisector of B A C

A B C = 1 2 b c sin A
A B D = 1 2 c A D sin A 2 and A D C = 1 2 b A D sin A 2
A B C = A B D + A D C
sin A = 2 sin A 2 cos A 2
As 0 < A < π , 0 < A 2 < π 2 cos A 2 > 0 cos A 2 = + 1 + cos A 2 as cos A = 2 cos 2 A 2 1
Use cos A = b 2 + c 2 a 2 2 b c and 2 s = a + b + c

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Let a , b , c represent the sides A , B , C respectively.

The angle bisectors are along a | a | + b | b | , b | b | + c | c | , c | c | + a | a |

Let the sides A B , B C , C A be x , y , z. Let A D be one of the angular bisector.
B D C D = x z
Hence
D = x c + z b x + z
What should be the next step? Or is there a better method?

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