Geometry with incircle and tangents.

Question

The points $M$ and $N$ are the points of tangency of the incircle of the isosceles triangle $\triangle ABC$ which are on the sides $AC$ and $BC$. The sides of equal length are $AC$ and $BC$. A tangent line $t$ is drawn to the minor arc $\overparen{MN}$. Suppose that $t$ intersects $AC$ and $BC$ at $Q$ and $P$ respectively. Suppose that the lines $AP$ and $BQ$ meet at $T$. Prove that $T$ lies on $MN$.

Answer 1

You can treat this as a limit case of Brianchon's theorem. Start with 6 points (green) on a circle (or in fact any conic):

Then you can draw tangents in these points (blue lines), form intersections of pairs of consecutive tangents (red points), and the three diagonals (red lines) will pass through one point (white). That's what the theorem says.

Now consider the limit case where $M_1\to M_2$ and $N_1\to N_2$. In the limit these are the same point, and the intersection of the corresponding tangents will be that same point as well. So all the $M_i$ are essentially the same point, and all the $N_i$ as well, so the line between $M_3$ and $N_3$ will be the line between $M$ and $N$, with $T$ on that line to show your claim. Also note that in the limit, $A,M,Q$ are collinear, as are $B,N,P$, forming two sides of your triangle intersecting in $C$.

Answer 2

I like coordinates. And I like the tangent half-angle formulas. So without loss of generality assume the incircle to be the unit circle, and assume $C$ to lie on the positive $x$ axis. Then you can write

$$M=\frac1{1+\alpha^2}\begin{pmatrix}1-\alpha^2\\2\alpha\end{pmatrix}\qquad N=\frac1{1+\alpha^2}\begin{pmatrix}1-\alpha^2\\-2\alpha\end{pmatrix}$$

as a generic pair of symmetric points, with some $\alpha\in(0,1)$. (Actually excluding $\alpha\in(-1,0)$ implies possibly relabeling points, again without loss of generality.) From this you get

$$A=\begin{pmatrix}-1\\\frac1\alpha\end{pmatrix}\qquad B=\begin{pmatrix}-1\\-\frac1\alpha\end{pmatrix}\qquad C=\begin{pmatrix}\frac{1+\alpha^2}{1-\alpha^2}\\0\end{pmatrix}$$

although you don't really need the coordinates of $C$. Now pick a point $R$ on the arc $MN$,as the contact point of tangent $t$:

$$R=\frac1{1+\beta^2}\begin{pmatrix}1-\beta^2\\2\beta\end{pmatrix} \qquad\text{with}\quad -\alpha<\beta<\alpha\;.$$

This leads to

$$Q=\frac1{1+\alpha\beta} \begin{pmatrix}1-\alpha\beta\\\alpha+\beta\end{pmatrix}\qquad P=\frac1{1-\alpha\beta} \begin{pmatrix}1+\alpha\beta\\-\alpha+\beta\end{pmatrix}$$

and intersecting the given lines you find

$$T=\frac1{1+\alpha^2}\begin{pmatrix}1-\alpha^2\\2\beta\end{pmatrix}$$

which has the same $x$ coordinate as $M$ and $N$, q.e.d.

I computed all of the above using homogeneous coordinates (thus the scale factor in front of most of these points, which would have been the third coordinate of the homogeneous coordinate vector). So connecting points and intersecting lines is simply computed as a cross product, while flipping the last coordinate turns a point on the unit circle into the tangent in that point and vice versa. I also used a bit of help from a computer algebra system, which was happy to deal with polynomials only thanks to the rational parametrization. This is not the most elegant proof to read, but it's a very powerful technique if you know there is a theorem and can't think of steps towards proving it.

Formally the proof could read like this. Define

\begin{align*} M&=[1-\alpha:2\alpha:1+\alpha]\\ N&=[1-\alpha:-2\alpha:1+\alpha]\\ R&=[1-\beta:2\beta:1+\beta]\\ c&=[1:0:1]\\ U&=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix} \end{align*}

w.l.o.g. then compute

\begin{align*} A&=(U\cdot M)\times c & B&=(U\cdot N)\times c \\ P&=(U\cdot N)\times(U\cdot R) & Q&=(U\cdot M)\times(U\cdot R)\\ T &= (A\times P)\times(B\times Q) & \det(M,N,T)&=0\qquad\text{q.e.d} \end{align*}

where $U$ is the matrix of the unit circle and the vanishing determinant checks for collinearity.