Showing two lines on a triangle coincide

Question

Let $M$ be the midpoint of (the smaller) arc $BC$ in circumcircle of triangle $ABC$. Suppose that the altitude drawn from $A$ intersects the circle at $N$. Draw two lines through circumcenter $O$ of $ABC$ paralell to $MB$ and $MC$, which intersect $AB$ and $AC$ at $K$ and $L$, respectively. Prove that $NK=NL$.

Could someone check my proof for this problem? At the end I'm doing something wrong but I can't find where I messed up.

Let the circumcircle of $\triangle ABC$ be the unit circle.

Without loss of generality, let $m=1$ (lower case letter denotes complex number corresponding to the point). Then it is easy to see $b,c$ are reflections across the real axis, so $bc=1$. Now we may use the parallel condition to get the equation $\frac{l}{c-m}\in\mathbb{R}\implies \frac{l}{c-m}=\frac{\overline{l}}{\overline{c-m}}$, and solving for $\overline{l}$, we get $\overline{l}=-\frac{l}{c}$. Now using the fact that $l\in AC$, we have $\frac{a-l}{l-c}\in\mathbb{R}\implies\frac{a-l}{l-c}=\frac{\overline{a-l}}{\overline{l-c}}$, and solving for $\overline{l}$, we get $\overline{l}=\frac{a+c-l}{ac}$. Equating the two and solving for $l$, we get $l=\frac{a+c}{1-a}$. Similarly, $k=\frac{a+b}{1-a}$.

I am almost positive the above is correct, so I think I am making the error in the next part of the proof:

Since $AN\perp BC$, then $AN$ is parallel to the real axis, so $an=-1$. Let $P$ be the midpoint of $KL$. We want to prove $KL\perp PN$ (i.e. $N$ lies on the perpendicular bisector of $KL$). To do this, we will prove $\frac{k-l}{p-n}\in i\mathbb{R}$, or $\frac{k-l}{p-n}+\frac{\overline{k-l}}{\overline{p-n}}=0$. Plugging in what we found earlier for $k,l,p$ (to get them in terms of $a,b,c$) and plugging in $n=-\frac{1}{a}$, we get that we want to prove: $$\frac{\frac{b-c}{1-a}}{\frac{2a+b+c}{1-a}+\frac{1}{a}}+\frac{\frac{\frac{c-b}{bc}}{\frac{a-1}{a}}}{\frac{\frac{2}{a}+\frac{1}{b}+\frac{1}{c}}{1-\frac{1}{a}}+a}=0$$

Which looks very intimidating, but we can easily simplify this to $$\frac{1}{\frac{2a+b+c}{1-a}+\frac{1}{a}}+\frac{a}{\frac{\frac{2}{a}+b+c}{1-\frac{1}{a}}+a}=0$$ by dividing both sides by $\frac{b-c}{a-1}$, and taking advantage of the fact that $bc=1$ (also then $\frac{1}{b}+\frac{1}{c}=b+c$).

But this simplifies to $a^2=1$, which is absurd since it would imply $M,N$ coincide, and therefore the triangle is isoceles. (I plugged this expression into my calculator, and that's what I got).

Where did I mess up? I checked all my steps and they all seem correct.

Answer 1

There are many mistakes in your writup of the first half of the proof, so I'll concentrate on the second half.

One can use a computer algebra system to check the computation. I'm using Sage.

sage: PR1.<a,b> = QQ[]  # Polynomial ring in two indeterminates
sage: def conj(x):      # Conjugation in its fraction field
...       return (x.numerator().subs(a=1/a, b=1/b) /
...               x.denominator().subs(a=1/a, b=1/b))
sage: c = 1/b
sage: l = (a+c)/(1-a)
sage: k = (a+b)/(1-a)
sage: p = (k+l)/2
sage: n = -1/a
sage: (k-l)/(p-n) + conj((k-l)/(p-n))
0
sage: res1 = ((b-c)/(1-a))/((2*a+b+c)/(1-a)+1/a) + (((c-b)/(b*c))/((a-1)/a))/((2/a+1/b+1/c)/(1-1/a)+a)
sage: res1
(a^3*b^3 - a^3*b - a*b^3 + a*b)/(-2*a^4*b^2 - 3*a^3*b^3 - a^2*b^4 + 3*a^3*b^2 + 2*a^2*b^3 - 3*a^3*b - 8*a^2*b^2 - 3*a*b^3 + 2*a^2*b + 3*a*b^2 - a^2 - 3*a*b - 2*b^2)
sage: factor(res1)
(-1) * b * (b - 1) * (b + 1) * a * (a - 1) * (a + 1) * (a^2*b + a*b^2 - a*b + a + 2*b)^-1 * (2*a^2*b + a*b^2 - a*b + a + b)^-1
sage: res2 = 1/((2*a+b+c)/(1-a)+1/a)+a/((2/a+b+c)/(1-1/a)+a)
sage: res2
(a^4*b^2 - a^3*b^2 - a^2*b^2 + a*b^2)/(2*a^4*b^2 + 3*a^3*b^3 + a^2*b^4 - 3*a^3*b^2 - 2*a^2*b^3 + 3*a^3*b + 8*a^2*b^2 + 3*a*b^3 - 2*a^2*b - 3*a*b^2 + a^2 + 3*a*b + 2*b^2)
sage: factor(res2)
a * (a + 1) * b^2 * (a - 1)^2 * (a^2*b + a*b^2 - a*b + a + 2*b)^-1 * (2*a^2*b + a*b^2 - a*b + a + b)^-1
sage: res1/res2 == (b - c)/(1 - a)
True

So yes, the term $\frac{k-l}{p-n}+\overline{\left(\frac{k-l}{p-n}\right)}$ is indeed equal to zero, as expected. But the first transformation you showed already has a different value. The subsequent simplification step is almost as you claimed in your post, except for an additional sign change. So that simplification in itself is valid, but the step before is wrong. So where exactly is the mistake?

sage: k - l == (b-c)/(1-a)
True
sage: p - n == (2*a+b+c)/(1-a)+1/a
False
sage: conj(k - l) == ((c-b)/(b*c))/((a-1)/a)
True
sage: conj(p - n) == (2/a+1/b+1/c)/(1-1/a)+a
False
sage: 2*p == (2*a+b+c)/(1-a)
True

Apparently you used $p=k+l$ but should have used $p=\frac{k+l}2$.

Answer 2

A solution through elementary geometry is also possible.

We have that $AM$ is the angle bisector of $\widehat{BAC}$, and by computing $\widehat{ALO}=\widehat{ACM}$ and $\widehat{AKO}=\widehat{ABM}$ it follows that $AKOL$ is a cyclic quadrilateral. Let $\Gamma_R$ be the circumcircle of $AKO$, $J=AN\cap\Gamma_R$ and $M'$ the antipode of $M$ in the circumcircle of $ABC$. Since $MAM'$ is a right triangle, we have that $\widehat{OKA}$ and $\widehat{AM'O}$ are supplementary angles, hence $M'\in\Gamma_R$, too, and $OJ\parallel MN$. From $NJ=OM'$ and $OJ\perp NM'$ it follows that $NJ=NO$.

On the other hand, it is not difficult to prove that $KL\perp NM'$, too, since $\widehat{KAJ}=\frac{\pi}{2}-\widehat{B}$ and $\widehat{OAL}=\widehat{BAC}-\widehat{BAO}=\widehat{KAJ}$. It follows that $NK=NL$ as wanted.

Moreover, it happens that the intersection of $NM'$ and $AM$ lies on $\Gamma_R$, too.