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? enter image description here

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    $\begingroup$ Can you made an image please? $\endgroup$ – Dr. Sonnhard Graubner May 10 '19 at 16:59
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    $\begingroup$ You might like the book "The Secrets of Triangles: A Mathematical Journey" by Alfred S. Posamentier and Ingmar Lehmann. It contains a figurative hundred $secrets$ I never dreamed about in triangles. $\endgroup$ – poetasis May 10 '19 at 17:06
  • $\begingroup$ @Dr.SonnhardGraubner image added $\endgroup$ – Zeno San May 10 '19 at 17:07

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 $\theta\to 0$, $EF\to 1-1+\dfrac{2(\frac{1}{2})^2}{(\frac32)(1)}\to\dfrac13$.

  • $\begingroup$ How $D=(\sin \theta,1-\cos \theta)$ ? $\endgroup$ – Zeno San May 10 '19 at 17:45
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    $\begingroup$ The $x$-coordinate is $AD\cdot \sin\theta$ and the $y$-coordinate is $AC-AD\cdot\cos\theta$. $\endgroup$ – CY Aries May 10 '19 at 17:47
  • $\begingroup$ i have h in terms of theta now, but how do i express EF in terms of theta ? @CY Aries $\endgroup$ – Zeno San May 10 '19 at 19:30

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