Mid-Midpoint

Suppose the homothety of ratio $2$ centered at $A$ sends $D$ to $D'$ and $M$ to $A'$ . Since it also sends $N$ to $O$ , the points $D',A',O$ are collinear. But $ABA'C$ is a parallelogram, so $CD' = AD' - AC = 2AD - AC = AB = CA',$ whence simple angle chasing yields $\angle{CA'D'} = \angle{BA'O} = 90^\circ-\frac{A}{2}$ .

Now extend ray $CA'$ past $A'$ to meet the circumcircle of $\triangle{ABC}$ at $E$ . As line $D'A'O$ bisects $\angle{BA'E}$ and $OB = OE$ , we must have $A'B = A'E$ . On the other hand, $\angle{BEA'} = 180^\circ - \angle{BAC} = \angle{BA'O}+\angle{CA'D'} = \angle{BA'E}$ gives $BA' = BE$ . Thus $\triangle{BA'E}$ is equilateral and so $180^\circ-A=60^\circ$ , or equivalently, $A = 120^\circ$ .

Let $MO$ meet circle $O$ at $P$ and $Q$ ( $Q$ and $A$ on same side of $BC$ ), $PE$ perpendicular to $AB$ . Then $E,M,D$ are projections of $P$ on three sides of $ABC$ [since $AD = \frac {AB+AC}{2}$ ]. Hence $E, M, D$ are collinear [Simson line], so $N$ also lies on that line. Since $AP$ perpendicular to $ED$ and $AP$ perpendicular to $AQ$ , $AQ \parallel NM$ . Since $N$ is the midpoint of $AO$ , $M$ is the midpoint of $OQ$ . Hence $A=120$ .

Correct me if I am wrong somewhere. I will be using special case of an right angled triangle. triangle3

Let's use coordinates. Let $AB$ have length $2$ and $AC$ have length $2c > 2$ . Let $A = (0,0), C = (2c, 0), B= (2\cos \theta, 2 \sin \theta)$ , where $90^\circ < \theta < 180^\circ$ . We have $M = (\cos \theta + c, \sin \theta)$ . By the length condition, we have $D = ( 1 + c, 0 )$ .

Let's determine $O$ using perpendicular bisectors. $AB$ has midpoint $(\cos \theta, \sin \theta)$ and slope $\frac {\sin \theta} {\cos \theta}$ . Hence, the perpendicular has slope $- \frac {\cos \theta}{\sin \theta}$ , so has equation $y = - \frac {\cos \theta}{ \sin \theta}x + \frac {1} {\sin \theta}$ . $AC$ has midpoint $(c,0)$ and perpendicular bisector $x = c$ . Hence, the center of the circle is $O = \left(c, \frac { 1 - c \cos \theta} {\sin \theta}\right)$ . This allows us to calculate that $N = \left(\frac {c}{2}, \frac { 1 - c \cos \theta} {2\sin \theta}\right)$ .

Now, since $M, N, D$ are a straight line, it follows that their slopes are equal. Hence,

$\ \frac {\sin \theta} {\cos \theta -1} = \mbox{slope of } MD = \mbox{slope of } ND = \frac { c \cos \theta - 1} {(c+2) \sin \theta}$

Expanding, we obtain $c ( \cos ^2 \theta - \sin ^2 \theta - \cos \theta) = 2 \sin^2 \theta + \cos \theta - 1$ , or that $0 = (c+1)(2 \cos^2 \theta - \cos \theta -1 ) = (c+1)(2 \cos \theta +1)(\cos \theta-1)$ . Hence, $\cos \theta = -\frac {1}{2}$ , so $\theta = 120 ^\circ$ .

construction: let the midpoint of AB be P. join M and P. extend MP to MF such that PF=AP=PB. Construct the angle bisector of angle BAC to intersect the circumcircle at D and BC at E. Extend MO to intersect circumcircle. it can be proven that MO intersects the circumcircle at D. extend MN to meet the angle bisector AD at J.

Proof: by Thales theorem we know that MP=AC/2 and by construction we know that PF=AB/2; hence MF=(AB+AC)/2. therefore AD=MF and AD parallel to MF. hence AFMD is parallelogram. =>/ BAC=/ APF; =>/ AFB=90-/ APF/2; =>/ AFB=90-/ BAC/2; =>/ ADM=/ AFB=/ ADJ=90-/ BAC/2;

consider triangle ADJ; / JAD=/ BAC/2; / ADJ=90-/ BAC/2; =>/ DJA=/ NJA=90;

consider: =>/ ODE=/ OAE=/ OAJ; =>/ JNA=/ ONM=90-/ JAN=90-/ OAE; =>/ JEM=/ OMN=90-/ JME=90-/ ODJ=90-/ OAE; =>/ ONM=/ OMN;

therefore in triangle OMN, / ONM=/ OMN; =>OM=ON=R/2 where R is circumradius in right triangle OMC OC=R &OM=R/2; =>/ MOC=60; =>/ COD=120; =>/ CAD=60; =>/ BAC=120;

HENCE /_BAC=120;