সমবাহুু ত্রিভুজ(বা $\pi/3$ কোণ) অঙ্কন, একটি প্রদত্ত বাহুর সাহায্যে

Alright, we have a compass, a straightedge, and a side.

Okay, now what? Well, let's call our first two vertices (the ones on the side we're already given) A and B, and let's draw AB.

Ok, what next?

Well, the other vertex is the same distance (the length AB) from point A, isn't it? So we can make a circle around point A (since a circle is all point a certain distance away from a center), and ONE of the points on that circle is vertex C (this is where the "They have 3 equal sides!" comes in handy).

So, here's our new diagram:

Well, what now? We have an infinite amount of points to choose from!

Hmm...think about this for a moment...

We do the same thing for point B!

In other words, we draw another circle around B!

The intersection points of the circles (let's call them Q and R) are where AQ and BQ and AR and BR are of the same length (and by the definition of a circle, this is the same length as AB!). Wait, 3 equal sides? That's exactly what we're looking for!

So, we've constructed two equilateral triangles (ABR and ABQ)!

Summary

In order to construct an equilateral triangle, you need to

1. Set your compass to the length of the side given

2. Construct a circle (or an arc, the point is to make sure that point C will be on that arc) of that length around the first point

3. Repeat step 2 with the second point

4 (optional). Draw a smiley face in the middle of the triangle (don't do this if you need to use the triangle :) )

Practice: Constructing an Equilateral Triangle

We want to construct an equilateral triangle using segment $\overline{AB}$.

Good job!

You're doing great!

Congrats!

You have constructed an equilateral triangle!

Explanation: An equilateral triangle is a triangle that is equilateral(yeah, no kidding), meaning that all of its sides are congruent to one another.

What we did in this construction was to use circles whose radii were all of length AB.

Because all radii of a circle are congruent, we have $\overline{AB}\cong\overline{AC}\cong\overline{BC}$, therefore allowing us to create equilateral $\triangle$ABC.

Hope this helps! :D

No Goal in Mind

Lili's post was really awesome, but there comes a time when you think, "How would a student discover goals by themselves?"

What object does it make sense to construct next?

What object does it make sense to construct next?

What would be a non-arbitrary circle?

We're going to assume you said a circle centered at $A$ containing $B$.

What circle?

What non-arbitrary things do you see?

Wee! We have two POINTS!

What do you notice?

What's a property of $C$?

How do you relate it to $A$ or $B$?

It's on the circle with center $A$ (or $B$) and radius $AB$.

What does this tell us?

$AC=AB$.

How else can you relate it to $A$ or $B$?

It's on the circle with center $B$ and radius $AB$.

What does this tell us?

$AB=BC$.

Do you notice anything?

$AB=BC=CA$. It is ...

Equilateral! Or you might be thinking of the $CA=CB$ thing, and you notice that it's ...

On the perpendicular bisector of $AB$!

The same can be said of $D$.