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The problem:

Let $M$ be the point of intersection between the diagonals of a cyclic quadrilateral $ABCD$, where $\angle AMB$ is acute. The isosceles triangle $BCK$, whose base is $BC$, is constructed externally to the quadrilateral, so that $\angle KBC$ + $\angle AMB$ = $90°$. Prove that $KM$ is perpendicular to $AD$.

I just have no idea how to do it. Sure $ABCD$ is cyclic so you can find a bunch of angles and do some angle chasing, but it got me nowhere.

If anyone could post a solution, or at least give me a hint on what to do, that would be appreciated. Thanks in advance.

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Elegant proof

Let's start with a generic cyclic quadrilateral, and define

\begin{align*} \angle ACB = \angle ADB &= \alpha \\ \angle CAD = \angle CBD &= \beta \\ \angle DBA = \angle DCA &= \gamma \\ \angle BAC = \angle BDC &= 180°-\alpha-\beta-\gamma \end{align*}

This answer is based on a key observation: $K$ is the center of the circle through $B,C,M$.

Illustration using circumcircle

This will certainly ensure that $BCK$ is isosceles with base $BC$. We'll check the angle sum condition along the way. If $K$ is the center of the circumcircle, then not only $BCK$ bit also $BMK$ and $CMK$ are isosceles, and we have the following conditions (introducing a new angle $\delta$ which we'll atempt to eliminate):

$$ \angle KBC=\angle BCK=\delta \\ \angle KMC=\angle MCK=\alpha+\delta \\ \angle BMK=\angle KBM=\beta+\delta \\ \angle CKM=180°-2(\alpha+\delta) \\ \angle MKB=180°-2(\beta+\delta) \\ \angle CKB=180°-2\delta= \bigl(180°-2(\alpha+\delta)\bigr)+\bigl(180°-2(\beta+\delta)\bigr) $$

From his last equation you can conclude

$$\delta=90°-\alpha-\beta$$

So now we can verify that

$$ \angle KBC+\angle AMB = \delta+(180°-\angle BAC-\angle DBA)\\ = 90°-\alpha-\beta+(180°-(180°-\alpha-\beta-\gamma)-\gamma)=90° $$

so $K$ is indeed the right point. Now consider the quadrilateral $ABKH$, where $H$ is the intersection of $KM$ with $AD$. It has interior angles

\begin{align*} \angle BAH &= \angle BAC + \angle CAD = (180°-\alpha-\beta-\gamma)+\beta = 180°-\alpha-\gamma \\ \angle KBA &= \angle KBC + \angle CBD + \angle DBA = (90°-\alpha-\beta) + \beta + \gamma = 90°-\alpha+\gamma \\ \angle HKB &= 180°-2(\beta+\delta) = 2\alpha \\ \angle AHK &= 360°-\angle BAH-\angle KBA-\angle HKB = 360°-180°+\alpha+\gamma-90°+\alpha-\gamma-2\alpha = 90° \end{align*}

$$\tag*{$\Box$}$$

Brute force proof

The above proof relies on spotting that fact about the circumcircle. Before I had that, I had a less elegant proof by brute force computation using sage. In fact, I found out that $BMK$ and $CMK$ are isosceles by comparing angles obtained from the construction below for random parameters. That computation is based on the following construction:

Construction

Start with four points on the unit circle, using homogeneous coordinates and a rational parametrization.

P.<a, b, c, d> = ZZ[]
def cpt(u): # a point on the unit circle
    return vector([1-u^2, 2*u, 1+u^2])
A = cpt(a)
B = cpt(b)
C = cpt(c)
D = cpt(d)

$$ A = \begin{pmatrix}1-a^2\\2a\\1+a^2\end{pmatrix} \qquad B = \begin{pmatrix}1-b^2\\2b\\1+b^2\end{pmatrix} \qquad C = \begin{pmatrix}1-c^2\\2c\\1+c^2\end{pmatrix} \qquad D = \begin{pmatrix}1-d^2\\2d\\1+d^2\end{pmatrix} $$

You can join points and intersect lines using the cross product.

def join(a, b):
    v = a.cross_product(b)
    g = gcd(v)
    return v/g
meet = join
M = meet(join(A, C), join(B, D))

$$ M=\begin{pmatrix} a b c - a b d + a c d - b c d - a + b - c + d \\ -2 a c + 2 b d \\ - a b c + a b d - a c d + b c d - a + b - c + d \end{pmatrix} $$

Next, construct the line through $M$ orthogonal to $BD$ and intersect this with $CD$ to obtain $E$. This will ensure $\angle AMB+\angle EMC=90°$.

perp = diagonal_matrix([1, 1, 0])
E = meet(join(perp*join(B, D), M), join(C, D))

$$ E=\begin{pmatrix} a b^{2} c^{2} d^{2} - a b^{2} c d^{3} + a b c^{2} d^{3} - b^{2} c^{2} d^{3} + a b^{2} c^{2} + a b^{2} c d - a b c^{2} d \\{}- b^{2} c^{2} d - 2 a b^{2} d^{2} + 2 a b c d^{2} - b^{2} c d^{2} + b c^{2} d^{2} - a b d^{3} + 2 b^{2} d^{3} \\{}+ a c d^{3} - 2 b c d^{3} - 2 a b c + b^{2} c + 2 a c^{2} - b c^{2} + a b d - a c d \\{}+ 2 b c d - 2 c^{2} d - a d^{2} - b d^{2} + c d^{2} + d^{3} - a + b - c + d \\[2ex] {}-2 a b c^{2} d^{2} + 2 b^{2} c d^{3} - 2 a b^{2} c + 2 a b^{2} d - 4 a b c d + 4 b^{2} c d + 2 a c^{2} d \\{}- 2 b c^{2} d + 2 a b d^{2} - 2 b^{2} d^{2} - 4 a c d^{2} + 4 b c d^{2} - 2 c^{2} d^{2} + 2 c d^{3} - 2 a c + 2 b d \\[2ex] {}- a b^{2} c^{2} d^{2} + a b^{2} c d^{3} - a b c^{2} d^{3} + b^{2} c^{2} d^{3} - a b^{2} c^{2} + a b^{2} c d - a b c^{2} d \\{}+ b^{2} c^{2} d - 2 a b c d^{2} + b^{2} c d^{2} - b c^{2} d^{2} + a b d^{3} - a c d^{3} + 2 b c d^{3} - 2 a b c \\{}+ b^{2} c - b c^{2} + a b d - a c d + 2 b c d - a d^{2} + b d^{2} - c d^{2} + d^{3} - a + b - c + d \end{pmatrix} $$

Compute $F$ as the other intersection of $BC$ with the circle through $CEF$. Due to the inscribed angle theorem, this will give you $\angle EFC$ = $\angle EMC$.

def cvec(p):
    x, y, z = p
    return vector([x*x + y*y, x*z, y*z, z*z])
P2.<mu> = P[]
p = list(matrix(map(cvec, [C, E, M, B+mu*C])).det())
F = p[1]*B - p[0]*C
F = F/gcd(F)

$$ F=\begin{pmatrix} {}- a b^{4} c^{2} + a b^{4} c d - a b^{3} c^{2} d + b^{4} c^{2} d - 2 a b^{3} c + b^{4} c + a b^{2} c^{2} - b^{3} c^{2} \\{}+ 3 a b^{3} d - 2 b^{4} d - 2 a b^{2} c d + 2 b^{3} c d - a b c^{2} d + b^{2} c^{2} d + a b^{2} - b^{3} + 2 a b c \\{}- 2 b^{2} c - 2 a c^{2} + 3 b c^{2} - a b d + b^{2} d + a c d - 2 b c d + a - b + c - d \\[2ex] 2 a b^{3} c^{2} - 2 b^{4} c d + 6 a b^{2} c - 4 b^{3} c - 2 a b c^{2} \\{}+ 4 b^{2} c^{2} - 4 a b^{2} d + 2 b^{3} d + 4 a b c d - 6 b^{2} c d + 2 a c - 2 b d \\[2ex] a b^{4} c^{2} - a b^{4} c d + a b^{3} c^{2} d - b^{4} c^{2} d + 2 a b^{3} c - b^{4} c + a b^{2} c^{2} \\{}+ b^{3} c^{2} - a b^{3} d - 2 b^{3} c d + a b c^{2} d - b^{2} c^{2} d + a b^{2} - b^{3} \\{}+ 2 a b c + b c^{2} - a b d - b^{2} d + a c d - 2 b c d + a - b + c - d \end{pmatrix} $$

Let $G$ be the midpoint of $BC$.

def midpoint(a, b):
    v = b[-1]*a + a[-1]*b
    g = gcd(v)
    return v/g
G = midpoint(B, C)

$$ G=\begin{pmatrix} - b^{2} c^{2} + 1 \\ b^{2} c + b c^{2} + b + c \\ b^{2} c^{2} + b^{2} + c^{2} + 1 \end{pmatrix} $$

Then you can construct the line parallel to $EF$ through $B$ and intersect this with the radius $OG$ to obtain $K$. The parallel line ensures $\angle KBC=\angle EFC$.

linf = vector([0, 0, 1])
O = vector([0, 0, 1])
K = meet(join(meet(join(E, F), linf), B), join(O, G))

$$ K=\begin{pmatrix} - a b c + b c d + a - d \\ a b + a c - b d - c d \\ a b c - a b d + a c d - b c d + a - b + c - d \end{pmatrix} $$

Now verify that the lines $KM$ and $AD$ are indeed orthogonal

join(K, M)[:2].dot_product(join(A, D)[:2]).is_zero()

This will print the desired answer, True.

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  • 1
    $\begingroup$ Quick proof that $K$ is the center of $\bigcirc BMC$: By the Inscribed Angle Theorem, all (and only) points $P$ separated from $K$ by $\overline{BC}$, such that $$\angle BPC = \frac{1}{2}(360^\circ - \angle BKC ) = \frac{1}{2}(360^\circ - (180^\circ - 2 (90^\circ - \angle AMB)) = 180^\circ - \angle AMB$$ lie on the (minor) arc $\stackrel{\frown}{BC}$ of the circle about $K$ through $B$ and $C$. $M$ is such a point. $\endgroup$ – Blue Apr 2 '14 at 20:34
  • $\begingroup$ Thank you! This problem has been haunting me for days, yet it can be solved so easily once you notice that little fact. $\endgroup$ – Deathkamp Drone Apr 3 '14 at 1:38

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