# Geometry problem based on triangles

Consider a right angled triangle $$ABC$$ , with right angle at $$C$$,$$ and $$|AC|=1$$. $$D$$ is a point on $$AB$$ such that $$|AD|=|AC|=1$$, and $$E$$ is a point on $$CB$$ such that $$, 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?

• Can you made an image please? – Dr. Sonnhard Graubner May 10 '19 at 16:59
• 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. – poetasis May 10 '19 at 17:06
• @Dr.SonnhardGraubner image added – 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 $$\theta\to 0$$, $$EF\to 1-1+\dfrac{2(\frac{1}{2})^2}{(\frac32)(1)}\to\dfrac13$$.

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