Consider a right angled triangle $ABC$ , with right angle at $C$,$ <CAB=\theta$ and $|AC|=1$. $D$ is a point on $AB$ such that $|AD|=|AC|=1$, and $E$ is a point on $CB$ such that $<CDE=\theta$, a perpendicular to $CB$ at $E$ is drawn which intersects $AB$ at $F$, Find $\lim_{\theta \to 0}|EF|$
How do i approach this problem?
Take $C$ as the origin and $A$ as the point $(0,1)$.
Then $B=(\tan\theta,0)$ and $D=(\sin\theta,1-\cos\theta)$
$\angle BCD=90^\circ-(180^\circ-\theta)/2=\theta/2$.
$\angle BED=\theta+\theta/2=3\theta/2$
Let $E=(h,0)$.
The slope of $DE$ is $\tan\dfrac{3\theta}{2}$.
\begin{align*} \frac{\cos\theta-1}{h-\sin\theta}&=\tan\frac{3\theta}{2}\\ h&=\sin\theta+\frac{\cos\theta-1}{\tan\theta\frac{3\theta}{2}} \end{align*}
As $\triangle ABC\sim\triangle FBE$, $\displaystyle \frac{\tan\theta-h}{EF}=\frac{\tan\theta}{1}$.
So, $\displaystyle EF=1-\frac{h}{\tan\theta}=1-\frac{\sin\theta}{\tan\theta}-\frac{\cos\theta-1}{\tan\frac{3\theta}{2}\tan\theta}=1-\cos\theta+\frac{\sin^2\frac{\theta}{2}\cos\frac{3\theta}{2}\cos\theta}{\sin\frac{3\theta}{2}\sin\theta}$
As $\theta\to 0$, $EF\to 1-1+\dfrac{2(\frac{1}{2})^2}{(\frac32)(1)}\to\dfrac13$.