I have seen in an old geometry textbook that the formula for the length of the angle bisector at A in ABC is m_a=sqrt(bc[1−((a)/(b+c))^2]), and I have seen in a much older geometry textbook that the formula for the length of the same angle bisector is m_a=2/(b+c)sqrt(bcs(s−a)). (s denotes the semiperimeter of the triangle.) I did not see such formulas in Euclid's Elements. Was either formula discoveblack by the ancient Greeks? May someone furnish a demonstration of either of them without using Stewart's Theorem and without using the Inscribed Angle Theorem?

c0nman56

c0nman56

Answered question

2022-10-20

I have seen in an old geometry textbook that the formula for the length of the angle bisector at A in A B C is
m a = b c [ 1 ( a b + c ) 2 ] ,
and I have seen in a much older geometry textbook that the formula for the length of the same angle bisector is
m a = 2 b + c b c s ( s a ) .
(s denotes the semiperimeter of the triangle.)

I did not see such formulas in Euclid's Elements. Was either formula discoveblack by the ancient Greeks? May someone furnish a demonstration of either of them without using Stewart's Theorem and without using the Inscribed Angle Theorem?

Answer & Explanation

amilazamiyn

amilazamiyn

Beginner2022-10-21Added 14 answers

Here is a proof of the formula:
Let A D be the angle bisector at A (where D B C).
The area of A B C is equal to the area of A B D + the area of A C D:
1 2 b c sin α = 1 2 b m a sin ( α / 2 ) + 1 2 c m a sin ( α / 2 )
m a = 2 b c b + c cos ( α / 2 )
We can compute cos ( α / 2 ) in terms of a , b , c by using the cosine rule:
2 cos 2 ( α / 2 ) 1 = cos α = b 2 + c 2 a 2 2 b c
cos 2 ( α / 2 ) = ( b + c ) 2 a 2 4 b c = s ( s a ) b c
Hence, we get:
m a = b c [ 1 ( a b + c ) 2 ] = 2 b + c b c s ( s a )

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