## 09 Dec Morley's Theorem

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**INTRODUCTION:**

One of the most elementary theorems in elementary geometry was discovered about 1899 by F. Morley (whose son Cristopher wrote novels such as *Thunder on the Left*). He mentioned it to his friends, who spread it over the world in the form of mathematical gossip. At last, after ten years, a trigonometrical proof by M. Satyanarayan and an elementary proof by M. T. Naraniengar were published.

**MORLEY’S THEOREM:**

*The three points of intersection of the adjacent triangles of the angles of any triangle form an equilateral triangle.*

In other words, any triangle ABC yields an equilateral triangle PQR if the angles A, B, C are trisected by AQ and AR, BR and BP, CP and CQ, as figure 1. (Much trouble is experienced if try a direct approach but the difficulties disappear if we work backwards, beginning with an equilateral triangle and building up a general triangle which is afterwards identified with the given triangle ABC.)

On the respective sides QR, RP, PQ of a given equilateral triangle PQR erect isosceles triangles P’QR, Q’RP, R’PQ whose base angles $$alpha, beta, gamma$$ satisfy the equation and the inequalities

$$alpha+beta+gamma = 120^{circ}, alpha leq 60^{circ}, beta leq 60^{circ}, gamma leq 60^{circ} .$$

Extend the sides of the isosceles triangles below their bases until they meet again in points A, B, C. Since $$alpha+beta+gamma+60^{circ}=180^{circ}, $$ we can immediately infer the measurements of some angles, as marked in figure 2. For instance, the triangle ABC is to describe it as lying on the bisector of the angle A at such a distance that

∠BIC = 90ᴼ + ½A.

Applying this principle to the point P in the triangle P’BC, we observe that the line PP’ (which is a median of both the equilateral triangle PQR and the isosceles triangle P’QR) bisects the angle at P’. Also the half angle at P’ is 90ᴼ – $$alpha$$ and

∠BPC = 180ᴼ – $$alpha$$ = 90ᴼ+ (90ᴼ – $$alpha$$ ).

Hence 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 three small angles at C are equal; likewise at A and at B. In other words, the angles of the triangle ABC are trisected.

The three small angles at A are each 1/3A = 60ᴼ – $$alpha$$ ; similarly at B and at C. Thus

$$alpha$$ = 60ᴼ – 1/3A, $$beta$$ = 60ᴼ – 1/3B, $$gamma$$ = 60ᴼ – 1/3C.

By choosing these values for the base angles of our isosceles triangles, we can ensure that the above procedures yield a triangle ABC that is similar to the given triangle.

This completes the proof.

**(SOURCE:** INTRODUCTION TO GEOMETRY, H.S.M. Coxeter, John Wiley & Sons, INC.)

Author:- Bishal Deb. Class – X. Email id: [email protected]

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Managing Editor of the English Section, Gonit Sora and Research Fellow, Faculty of Mathematics, University of Vienna.

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