How to prove by vector method that internal angle bisectors of a triangle are concurrent ?
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Let A(a), B(b), C(c) be the position vectors of the vertices of the triangle ABC and the lengths of sides BC, CA and AB l,m,n respectively.
The Internal bisectors of the triangle divide the opposite sides in the ratio of the sides containing the angles.
Since AD is the internal bisector of the 1
Position vector of D = nc+mb/m+n
let the Internal bisectors intersect at I
ID/AI = BD/AB -->2
BD/DC=m/n (using 1)
CD/BD = m/n
=> CD+BD/BD = m+n/n
=> BC/CD = m+n/n
=> BD = ln/m+n -->3
From (2) and (3) we get ,
ID/AI = ln/(m+n)n = l/m+n
Position vector of I = [nc+mb/m+n](m+n)+la/l+m+n = la+mb+nc/l+m+n
Similarly , we can prove that I lie on the internal bisectors of angles B and C.
The Internal bisectors of the triangle divide the opposite sides in the ratio of the sides containing the angles.
Since AD is the internal bisector of the 1
Position vector of D = nc+mb/m+n
let the Internal bisectors intersect at I
ID/AI = BD/AB -->2
BD/DC=m/n (using 1)
CD/BD = m/n
=> CD+BD/BD = m+n/n
=> BC/CD = m+n/n
=> BD = ln/m+n -->3
From (2) and (3) we get ,
ID/AI = ln/(m+n)n = l/m+n
Position vector of I = [nc+mb/m+n](m+n)+la/l+m+n = la+mb+nc/l+m+n
Similarly , we can prove that I lie on the internal bisectors of angles B and C.
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