Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's consider the right triangle ABC with angle $\angle ACB = \frac{\pi}{2}$, $\angle{BAC}=\alpha$, D is a point on AB such that |AC|=|AD|=1; the point E is chosen on BC such that $\angle CDE=\alpha$. The perpendicular to BC at E meets AB at F. I'm supposed to calculate:

$$\lim_{\alpha\rightarrow0} |EF|$$

enter image description here

share|cite|improve this question
How about a picture to see what's going on? – Sean Jun 11 '12 at 13:22
up vote 7 down vote accepted

Below is the construction described in the question:

$\hspace{3.5cm}$triangle geometry

The Inscribed Angle Theorem says that $\angle DCB=\angle DCE=\alpha/2$. Since $\angle CDE=\alpha$ by construction, the exterior angle $\angle DEB=3\alpha/2$.

Since $\angle DBE=\pi/2-\alpha$, as $\alpha\to0$, $\triangle DBE$ and $\triangle DBC$ get closer to right triangles with $$ \frac{|BE|}{|DB|}\tan(3\alpha/2)\to1\quad\text{and}\quad\frac{|BC|}{|DB|}\tan(\alpha/2)\to1 $$ Thus, $$ \frac{|BE|}{|BC|}\frac{\tan(3\alpha/2)}{\tan(\alpha/2)}\to1 $$ Therefore, by similar triangles and because $\lim\limits_{x\to0}\frac{\tan(x)}{x}=1$, $$ \frac{|FE|}{|AC|}=\frac{|BE|}{|BC|}\to\frac13 $$ Since $|AC|=1$, we have $$ \lim_{\alpha\to0}|FE|=\frac13 $$

share|cite|improve this answer
nice approach and very simple. Thank you! – user 1618033 Jun 11 '12 at 14:42
+1 elegant and so simple understandable way. – Babak S. Jun 11 '12 at 17:52

We do a computation similar to the one by robjohn. Let $\alpha=2\theta$. By angle-chasing, we can express all the angles in the picture in terms of $\theta$.

Since $\triangle ACD$ is isosceles, $CD=2\sin\theta$. Now we compute $DE$ by using the Sine Law. Note that $\angle CED=\pi-3\theta$ and $\angle DCE=\theta$. Thus $$\frac{DE}{\sin\theta}=\frac{CD}{\sin(\pi-3\theta)}=\frac{CD}{\sin 3\theta},$$ and therefore $$DE=\frac{2\sin^2\theta}{\sin 3\theta}.$$ Now use the Sine Law on $\triangle DEF$. We get $$\frac{EF}{\sin(\pi/2+\theta)}=\frac{DE}{\sin 2\theta}.$$ Since $\sin(\pi/2+\theta)=\cos\theta$, we get $$EF=\frac{2\sin^2\theta\cos\theta}{\sin 2\theta\sin 3\theta}.$$ Using the identity $\sin 3\theta=3\sin\theta-4\sin^3\theta$, we find the very simple result $$EF=\frac{1}{3-4\sin^2\theta},$$ and therefore the limit is $1/3$.

share|cite|improve this answer
Very nice and short! – user 1618033 Jun 11 '12 at 19:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.