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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

1171.0. "proof for pretty theorem in 20.3?" by COOKIE::MURALI () Fri Dec 29 1989 13:29

    Hello,
    
    I am going way back to 20.3. Does anybody have a proof for that
    theorem. It would be nice to have a proof that is based only
    on theorems in geometry but any trignometric proof would also do.
    
    Thanks,
    
    Murali.

T.R Title User Personal
Name Date Lines

1171.1 Morely's Theorem SNOFS1::MOHAN Sun Jan 21 1990 22:21 56

T.R	Title	User	Personal Name	Date	Lines
1171.1	Morely's Theorem	SNOFS1::MOHAN		`Sun Jan 21 1990 22:21`	56
	Morely's Theorem ================ The proof of this theorem has been taken from H.S.M Coexter's 'Introduction To Geometry'. The accompanying figure is too difficult to draw on the terminal. So I am leaving it out. The theorem states that "The 3 points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle". Morely discovered this proof in 1899. But it was only in 1909 that it was proved by two Indian mathematicians M.Satyanarayanan (trigonometrical proof) and M.T.Naraniengar (elemenatry geometrical proof). If you try a direct approach to this proof it is very difficult. The trick is to work backwards with an equilateral triangle and building up a general triangle which is afterwards identified with the given triangle ABC. Triangle ABC yields an equil.triangle PQR if the angles A,B,C are trisected by AQ, AR, BP, BR, CP and CQ. On the respective sides QR, RP, PQ of the given equ.triangle PQR, erect isosceles triangles P'QR, Q'RP, R'PQ whose base angles a, b, c satisfy the equation a + b + c = 120, a < 60, , b < 60, c < 60. Extend the sides of the isosceles triangles below their bases until they meet again in points A, B, C. Since a + b + c + 60 = 180, <AQR = 60 - a. Now one way to characterize the incenter I of triangle ABC is to describe it as lying on the bisector of the angle A at such a distance that <BIC = 90 + A/2. Applying this principle to the point P in the triangle P'BC we observe that the line PP' which is the median of both the equ.triangle PQR and the isosceles triangle P'QR bisects the angle at P'. Also the half angle at P' is 90 - a and <PBC = 180 - a = 90 + (90 - a) Therefore P is the incenter of the triangle P'BC. Likewise Q is the incenter of Q'CA and R of R'AB. Therefore all the 3 small angles at C are equal, likewise at A and at B. In otherwords the angles of triangle ABC are trisected. The 3 small angles at A are each A/3 = 60 - a. Similarly at B and C. Thus a = 60 - A/3, b = 60 - B/3, c = 60 - C/3. By choosing these values for the base angles of out isosceles triangle we can ensure that the above procedure yields a triangle ABC that is similar to any given triangle. Q.E.D Regards, Raj

    
			Morely's Theorem
			================
The proof of this theorem has been taken from H.S.M Coexter's 
'Introduction To Geometry'.

The accompanying figure is too difficult to draw on the terminal. So I am
leaving it out.

The theorem states that "The 3 points of intersection of the adjacent
trisectors of the angles of any triangle form an equilateral triangle".

Morely discovered this proof in 1899. But it was only in 1909 that it was
proved by two Indian mathematicians M.Satyanarayanan (trigonometrical proof)
and M.T.Naraniengar (elemenatry geometrical proof).

If you try a direct approach to this proof it is very difficult. The trick
is to work backwards with an equilateral triangle and building up a general
triangle which is afterwards identified with the given triangle ABC.

Triangle ABC yields an equil.triangle PQR if the angles A,B,C are trisected
by AQ, AR, BP, BR, CP and CQ.

On the respective sides QR, RP, PQ of the given equ.triangle PQR, erect
isosceles triangles P'QR, Q'RP, R'PQ whose base angles a, b, c satisfy the
equation a + b + c = 120, a < 60, , b < 60, c < 60.

Extend the sides of the isosceles triangles below their bases until they meet
again in points A, B, C.

Since a + b + c + 60 = 180, <AQR = 60 - a.
Now one way to characterize the incenter I of triangle ABC is to describe it
as lying on the bisector of the angle A at such a distance that
<BIC = 90 + A/2.

Applying this principle to the point P in the triangle P'BC we observe that
the line PP' which is the median of both the equ.triangle PQR and the
isosceles triangle P'QR bisects the angle at P'.

Also the half angle at P' is 90 - a and <PBC = 180 - a = 90 + (90 - a)
Therefore P is the incenter of the triangle P'BC. Likewise Q is the incenter
of Q'CA and R of R'AB. Therefore all the 3 small angles at C are equal,
likewise at A and at B. In otherwords the angles of triangle ABC are
trisected.
The 3 small angles at A are each A/3 = 60 - a. Similarly at B and C.
Thus a = 60 - A/3, b = 60 - B/3, c = 60 - C/3.

By choosing these values for the base angles of out isosceles triangle
we can ensure that the above procedure yields a triangle ABC that is similar
to any given triangle.

				Q.E.D

Regards,
Raj