# An alternative proof of 30-60-90 theorem/

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.

• What is the 30-60-90 theorem? Are you talking about the proportions of the side lengths? The equilateral triangle is pretty easy... – apnorton Aug 28 '13 at 13:32
• It is a triangle with angles 30, 60 and 90 degrees. The theorem states that side opposite to 30 degree angle is half of the hypotenuse. – user61681 Aug 28 '13 at 14:01
• I just want some simple proof or explanation of the fact that side opposite to 30 degree angle is half of the hypotenuse. I don't want explanation for the angle opposite to 60 degree. – user61681 Aug 28 '13 at 14:04
• @user61681: 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! – Blue Aug 28 '13 at 14:41
• If you want an alternative proof, it would be useful to show us the proof you already have. – robjohn Aug 28 '13 at 17:01

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

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!

• +1 I'm wondering if it will work better to ask them about the sides of the equilateral triangle first, then fold it and ask them about the ratio of sides, then ask about the angles of the triangle. – Calvin Lin Aug 28 '13 at 16:19

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.

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, $$b=2c$$

• The answer is okay. But I want explanation suitable at school level, possibly only using construction,congruence of triangles, Pythagoras theorem... – user61681 Aug 28 '13 at 14:25
• @user61681 i have edited it, will it do, reply if you want anything else – Shobhit Aug 28 '13 at 14:29
• @Shobhit: Surely the equilateral triangle construction comes conceptually before knowing the value of $\sin 30^\circ$ (or the Law of Sines), doesn't it? (You wouldn't argue that the sides of a square are equal because $\tan 45^\circ = 1$, would you?) – Blue Aug 28 '13 at 14:34
• The students don't know trigonometry. May be the answer using area of triangle/rectangle, Pythagoras theorem, congruence of triangles...etc – user61681 Aug 28 '13 at 14:34

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.

• This proof already known. I need an alternate and equally simple proof. – user61681 Aug 28 '13 at 15:26
• ok, will think about it but without trig and without the use of an equiliateral triangle, it is going to be difficult I think – imranfat Aug 28 '13 at 15:45
• Thanks for your reply and interest. – user61681 Aug 28 '13 at 15:46

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.

• Thanks for your nice and simple answer. – user61681 Aug 29 '13 at 15:20
• @user61681 The advantage is not going "outside" the original triangle, but it has three congruent triangles instead of the two involved when bisecting an equilateral triangle. A toss-up which is "simpler"... – coffeemath Aug 29 '13 at 17:06

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 $$\sin 30^{\circ}=\frac 12$$

Its a very easy proof indeed. It does involve trigonometric ratio but the ratio is limited at only one value- sin. Infact it can also be proved by cos,tan,etc. Proof We know sinx=opp/hypotenuse. Lets consider a right angled triangle ABC, in which AC is hypotenuse, and angle B=90º angle A=30º and angle C=60º.

Hence sin30=BC/AC 1/2= BC/AC Hence 1/2 BC= AC