In a triangle ABC, a circle is inscribed with center in $I$. The inscribed circle touches sides $BC,CA,AB$ in $D,E,F$ respectively. Join the point $C$ and $F$, $B$ and $E$. Let $Q$ and $R$ be the point of intersection of the segments $BE$ and $CF$ with the inscribed circle respectively. Let $M$ be the intersection point between the side $BC$ and the line that pases through $ER$.

Show that $ MF, DE, QR$ are concurent

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    $\begingroup$ What is the question? Are you supposed to show that those three lines intersect at a single point? $\endgroup$ – abiessu Oct 1 '13 at 15:58
  • $\begingroup$ Hello,$MF,DE,QR$ $\endgroup$ – china math Oct 1 '13 at 16:05
  • $\begingroup$ Some observations: you can repeat the whole process using permuted labels for the corners of the triangle. In that way, you can obtain a total of 6 such points of concurrence. They lie on a common conic but not on a circle. Each edge of the Gergonne triangle contains two such points, which together with the endpoints of that edge form a fixed cross ratio, namely $\frac43$, like equidistant points. See this illustration. Not sure whether any of this will help, probably not. Interesting nevertheless, I think. $\endgroup$ – MvG Oct 1 '13 at 21:53

I had hoped for a more beautiful solution, but since none came up, here is an algebraic solution. This will be expressed in homogenous coordinates, so connecting points and intersecting lines can both be computed using the cross product.

Assume the incircle is the unit circle. Then you can choose three parameters $\theta_D,\theta_E,\theta_F$ to describe $D,E,F$ like this:

\begin{align*} D &= \begin{pmatrix}\cos\theta_D\\\sin\theta_D\\1\end{pmatrix} & E &= \begin{pmatrix}\cos\theta_E\\\sin\theta_E\\1\end{pmatrix} & F &= \begin{pmatrix}\cos\theta_F\\\sin\theta_F\\1\end{pmatrix} \end{align*}

You also need the matrix of the circle itself:

\begin{align*} M &= \begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix} \end{align*}

From this, you can compute everything else as follows:

\begin{align*} a &= M\cdot D & b &= M\cdot E & c &= M\cdot F \\ A &= b\times c & B &= a\times c & C &= a\times b \end{align*} \begin{align*} Q &= \left(2B^TME\right)B - \left(B^TMB\right)E \\ R &= \left(2C^TMF\right)C - \left(C^TMC\right)F \\ M &= (E\times R)\times a \end{align*} \begin{align*} \ell_{FM} &= F\times M & \ell_{DE} &= D\times E & \ell_{QR} &= Q\times R \end{align*}

At the end of the day, you have three vectors for the three lines. To check that they are concurrent, write them as the columns of a single matrix, then compute its determinant. That determinant will be zero iff the lines are concurrent.

You can use a computer algebra system of your choice to show that this final expression is indeed zero, independent of the three variable angles.

Sage code to do this:

var('thetaD', domain='real', latex_name='\theta_D')
var('thetaE', domain='real', latex_name='\theta_E')
var('thetaF', domain='real', latex_name='\theta_F')
D = vector([cos(thetaD), sin(thetaD), 1])
E = vector([cos(thetaE), sin(thetaE), 1])
F = vector([cos(thetaF), sin(thetaF), 1])
M = diagonal_matrix(QQ, [1, 1, -1])
a = M*D
b = M*E
c = M*F
A = b.cross_product(c)
B = a.cross_product(c)
C = a.cross_product(b)
Q = 2*(B.row()*M*E.column())[0,0]*B - (B.row()*M*B.column())[0,0]*E
R = 2*(C.row()*M*F.column())[0,0]*C - (C.row()*M*C.column())[0,0]*F
M = E.cross_product(R).cross_product(a)
FM = F.cross_product(M)
DE = D.cross_product(E)
QR = Q.cross_product(R)
Matrix([FM, DE, QR]).det().is_zero()

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