Limit of ratio of areas of triangles defined by tangents to a circle Let $AB $ be an arc of a circle. Tangents are drawn at $A $ and $B $ to meet at $C $. Let $M $ be the midpoint of arc $AB $. Tangent drawn at $M $ meet $AC $ and $BC $ at $D $, $E $ respectively. Evaluate $$\lim_{AB \to 0}\frac {\Delta ABC}{\Delta DEC} $$ I don't think evaluating the limit will be a problem. But how do I find the areas of the triangles and in what parameters?
 A: This is plain geometry: join the point $\;C\;$ with the circle's center $\;O\;$ , say. Then, $\;M\;$ is on $\;OC\;$ (why?) and furthermore, $\;OC\perp AB\;,\;\;OC\perp DE\;$ ( why? Show that $\;AB\parallel DE\;$), which in fact means that in fact $\;\Delta ABC\sim\Delta DEC\;$ .
Now, let us observe that since $\;|CM|=|CO|-R\;,\;\;R=$ the circle's radius, whereas if we denote by $\;P\;$ the intersection point of $\;AB\;$ with $\;OC\;$ , using Pythagoras theorem we get
$$|OP|^2=R^2-\left(\frac12|AB|\right)^2=R^2-\frac14|AB|^2$$
so 
$$|CP|=|CO|-|OP|=|CO|-\frac12\sqrt{4R^2-|AB|^2}$$
and the similarity ratio is then:
$$\frac{|CM|}{|CP|}=\frac{|CO|-R}{|CO|-\frac12\sqrt{4R^2-|AB|^2}}$$
Well, now just use that $\;\frac{\Delta ABC}{\Delta DEC}=\left(\frac{|CM|}{|CP|}\right)^2\;$ and observe that when $\;AB\to 0\;$, all the quantities involved are constant except, of course $\;|AB|\;$ , which also tends to zero.
A: Hint: draw a circle of radius $1$ around the origin. You can assume that $AB$ is symmetrical w.r.t. $Y$-axis. Let $A=(\cos(\alpha), \sin(\alpha))$ be in the first quadrant, $B=(-\cos(\alpha), \sin(\alpha))$ with $ 0 \lt \alpha \leq \frac{1}{2}\pi$. Then $|\Delta ABC|=\cos(\alpha)\sin(\alpha)$ and $|\Delta DEC|=\frac{(2\sin(\alpha)-1)^2}{\tan(\alpha)}$. Now take your limit where $\alpha \rightarrow \frac{1}{2}\pi$, which should give you $1$. Generalize this to a circle with radius $r$.
A: Let $A(r\cos\theta,r\sin\theta),B(r\cos\phi,r\sin\phi)\in \{(x,y)\mid x^2+y^2=r^2\}$ then: $$C\left(r\frac{\cos\left(\frac{\phi+\theta}2\right)}{\cos\left(\frac{\phi-\theta}2\right)},r\frac{\sin\left(\frac{\phi+\theta}2\right)}{\cos\left(\frac{\theta-\phi}2\right)}\right)\\
M\left(r\cos\left(\frac{\theta+\phi}2\right),r\sin\left(\frac{\theta+\phi}2\right)\right)\\
$$
And:
$${\rm AB}:x\cos\left(\frac{\theta+\phi}2\right)+y\sin\left(\frac{\theta+\phi}2\right)-r\cos\left(\frac{\theta-\phi}2\right)=0\\
{\rm DME}:x\cos\left(\frac{\theta+\phi}2\right)+y\sin\left(\frac{\theta+\phi}2\right)-r=0$$
So:
$$\frac {\Delta {\rm ABC}}{\Delta {\rm DEC}}
=\left|\frac{\frac{\cos ^2\left(\frac{\theta +\phi }{2}\right)}{\cos \left(\frac{\theta -\phi }{2}\right)}+\frac{\sin ^2\left(\frac{\theta +\phi }{2}\right)}{\cos \left(\frac{\theta -\phi }{2}\right)}-\cos \left(\frac{\theta -\phi }{2}\right)}{\frac{\cos ^2\left(\frac{\theta +\phi }{2}\right)}{\cos \left(\frac{\theta -\phi }{2}\right)}+\frac{\sin ^2\left(\frac{\theta +\phi }{2}\right)}{\cos \left(\frac{\theta -\phi }{2}\right)}-1}\right|^2=4\cos^4\frac{\theta-\phi}4$$
Now as ${\rm AB}\to0$ or $\theta-\phi\to0$
$$\huge\lim_{AB \to 0}\frac {\Delta {\rm ABC}}{\Delta {\rm DEC}}=4$$
A: Without loss of generality, let the radius of the circle be 1 and the center of the circle be at (0,0). Moreover let the coordinates of $A$ and $B$ be $(\cos\theta,\sin\theta)$ and $(\cos\theta,-\sin\theta)$, respectively. Then the coordinates of $M$ and $C$ are (1,0) and $(\sec\theta,0)$, respectively. Let the intersection of $AB$ and $OC$ be $N$. Thus
\begin{eqnarray}
|NC|=|OC|-|ON|=\sec\theta-\cos\theta,|MC|=\sec\theta-1.
\end{eqnarray}
Note that $\Delta ABC$ and $\Delta DEC$ are similar and hence
\begin{eqnarray}
\lim_{|AB|\to0}\frac{S_{\Delta ABC}}{S_{\Delta DEC}}&=&\lim_{|AB|\to0}\left(\frac{|NC|}{|MC|}\right)^2\\
&=&\lim_{\theta\to0}\left(\frac{\sec\theta-\cos\theta}{\sec\theta-1}\right)^2\\
&=&\lim_{\theta\to0}(1+\cos\theta)^2\\
&=&4.
\end{eqnarray}
