An alternative proof of 30-60-90 theorem/

Question

A 30-60-90 theorem in Geometry is well known. The theorem states that, in a 30-60-90 right triangle, the side opposite to 30 degree angle is half of the hypotenuse

I have a proof that uses construction of equilateral triangle. Is the simpler alternative proof possible using school level Geometry. I want to give illustration in class room.

Answer 1

Consider the following picture,

Applying the law of sines (http://en.wikipedia.org/wiki/Law_of_sines) in triangle ABC, we have

$$\frac{a}{\sin 60}=\frac{b}{\sin 90}=\frac{c}{\sin 30}$$

or $$\frac{b}{1}=\frac{c}{\frac1{2}}$$

or $$b=2c$$

EDIT

Consider triangle ABC, we have

$$\sin 30 = \frac{c}{b}$$

or equivalently $$\frac1{2}=\frac c{b}$$

or, $$b=2c$$

Answer 2

If you take 4 identical triangles like that, you can easily arrange them into an rectangle such that:

Two short sides rest on the hypotenuse of another such that the 90 degree angles touch.

The remaining triangle's 60 degree angle touch where two other 60 degree angles touch with the hypotenuse against that of another triangle

As I assume they know how to calculate area, angles, and $a^2+b^2=c^2$, they should be able to play with this figure to show it is a perfect rectangle and all sides and areas are correct.

Answer 3

Converting a comment to an answer, as suggested ...

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Prep by folding a cardboard equilateral triangle in half. Show the audience the $30^\circ$-$60^\circ$-$90^\circ$ right triangle, then unfold to show the equilateral.

Ta-dah!

Answer 4

Draw an equilateral triangle. In one vertex you draw a perpendicular that intersects the extension of the opposide side of the vertex from which you drew the perpendicular.

You now have a $30-60-90$ triangle (can you see why?) from which you can derive the $1-2-\sqrt{3}$ ratios. No trig needed.

Answer 5

In right triangle $ABC$ with the right angle at $B$ and the $30$ degree angle at $A$, bisect the $60$ degree angle at $C$, and let the point on segment $AB$ where this bisector crosses be $D$. Now drop a perpendicular from $D$ to meet segment $AC$ at $E$.

We now have three smaller $30-60-90$ triangles which are all congruent to each other. Giving each with the ordering "vertex angle 30, vertex angle 90, vertex angle 60", these are $$\Delta AED, \ \Delta CED, \ \Delta CBD.$$ The angles match so they are similar, while the first two share side $ED$ and the second two share side $CD$, so they are in fact congruent.

Now noting that $AC=AE+EC,$ we have $$\frac{AC}{BC}=\frac{AE}{BC}+\frac{EC}{BC}=1+1=2,$$ using from corresponding parts of congruent triangles that $AE=CB=BC$ and $EC=CE=CB=BC.$

I think this argument uses only congruent triangles, and the (fairly simple) ideas of angle bisectors and perpendiculars.

Answer 6

It seems to me that any proof has to work with the specific properties of the angles involved - like three angles of $60^{\circ}$ make a straight line, and that $2\times 30=60$ - the easiest way to use this information is to divide an equilateral triangle in two, and essentially every proof you get will either use that fact or conceal it.

Here is a short proof using a double angle formula and that $\sin A=\cos (90^{\circ}-A)$

$$\sin 60^{\circ}=2\sin 30^{\circ}\cos30^{\circ}=2\sin 30^{\circ}\sin60^{\circ}$$ whence $$\sin 30^{\circ}=\frac 12$$