Proof of Pythagorean Theorem

PowerOfPi

Does anyone know how to prove the Pythagorean Theorem?

I use it every day but I don't know how it works.

**Posted:** Fri May 08, 2009 4:56 pm

RoFlLoLcOpT

You could use trigonometry to prove it.

trig proof

geometric proof

In addition to these proofs, there are MANY MANY more. Try to find some of them by yourself!

**Posted:** Fri May 08, 2009 5:12 pm

PowerOfPi

Thank you. That was really helpful and now I can understand it.

**Posted:** Fri May 08, 2009 5:16 pm

winsome · New Member Offline Joined: 07 Dec 2008 Posts: 7

To learn more you may click the link below and go through video lesson on ‘Pythagorean Theorem'.

http://www.winpossible.com/lessons/Geometry_Getting_Started_-_Pythagorean_theorem.html

**Posted:** Sun May 10, 2009 8:39 pm

Bachukas

**Posted:** Mon May 11, 2009 11:24 am

PowerOfPi

Good point.

**Posted:** Mon May 11, 2009 1:33 pm

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AceOfDiamonds

dot("A", (0,0));
dot("B", (0,3));
dot("C", (4,0));
dot("M", (1.44,1.92));
draw((0,0)-...

Ok. $M$ is projection of $A$ on $BC$ .
$\angle CAB \cong \angle AMB$
$\angle ABC \cong \angle MBA$
You get $\triangle ABC \sim \triangle MBA$
And from that you get $\frac {CB}{AB} = \frac {AB}{MB}$ and from that
$AB^2 = MB\times BC$
Similarly, you do the same stuff and you get
$AC^2 = CM\times CB$
Now add the two equations and you get
$AB^2 + AC^2 = MB\times BC + CM\times CB$
now since $CB$ is $BC$ , you get
$AB^2 + AC^2 = MB(BC) + CM(BC)$
now factoring out $BC$ gives
$AB^2 + AC^2 = (MB + CM)(BC)$
Now since $MB + CM = BC$ , you get
$\boxed{AB^2 + AC^2 = BC^2}$
$Q.E.D$ Smile

(First time using asymptote. It takes a long time to type. Also, I came up with this proof using Alt. on Hyp. theorems myself. Not that I'm saying I'm the first or anything)

Mukherjee · Poincare Conjecture Offline Joined: 13 Sep 2008 Posts: 141

**Posted:** Tue May 12, 2009 1:07 am

akhil0422

How is $AB^2 + AC^2 = BC^2 = A^2 + b^2 = C^2$ .

Or am I just being stupid?

And what is QED? You and RofLoc... whatever... use it all the time.

**Posted:** Tue May 12, 2009 3:39 pm

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RoFlLoLcOpT

google is your friend

you're being stupid Razz

jk

This is the pythagorean theorem...

$\text{leg}_1^2+\text{leg}_2^2=\text{hypotenuse}^2$ . In this case, $\text{leg}_1=AB, \text{leg}_2=AC, \text{hypotenuse}=BC$

**Posted:** Tue May 12, 2009 3:46 pm

AceOfDiamonds

**Posted:** Tue May 12, 2009 4:27 pm

u63rn3rd · New Member Offline Joined: 01 Mar 2009 Posts: 6

I think this is a good explanation.

**Posted:** Tue May 12, 2009 5:57 pm

u63rn3rd · New Member Offline Joined: 01 Mar 2009 Posts: 6

oops, I accidentally cut off the equations. they say 5^2+12^2=169 and 169 ^.5 = 13 Embarassed

**Posted:** Tue May 12, 2009 6:12 pm

PowerOfPi

It is a good explantation but it only proves the 5-12-13 triangle.

**Posted:** Tue May 12, 2009 6:37 pm

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AceOfDiamonds

ummm that 'explanation' doesn't really explain anything. 5-12-13 is a well-known pythagorean triple, but that applet doesn't show anything about the pythagorean theorem other than an example. It doesn't even prove that the triangle is a right triangle.

**Posted:** Tue May 12, 2009 9:19 pm

nick42 · Yang-Mills Theory Offline Joined: 10 Jun 2008 Posts: 533

**Posted:** Wed May 13, 2009 12:05 pm

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RoFlLoLcOpT

I'm pretty sure everyone knows the square proof...

When you prove something, you can't use an example. The Pythagorean Theorem must be proved for all numbers $a, b, c,$ not just certain example lengths.

**Posted:** Wed May 13, 2009 12:55 pm

u63rn3rd · New Member Offline Joined: 01 Mar 2009 Posts: 6

sorry I thought you were just looking for the example. here is a great NLVM manipulative that explains the Pythagorean theorem:
http://nlvm.usu.edu/en/NAV/frames_asid_164_g_4_t_3.html?open=instructions&from=category_g_4_t_3.html
Embarassed

**Posted:** Wed May 13, 2009 2:33 pm

HiDN428

one question. why cant the theorem $[sin^2]$ + $[cos^2]$ = $1$ be a proof of the pythag theorem. i guess the pythag proves that theorem but it works vice versa as well

**Posted:** Wed May 13, 2009 2:34 pm

AceOfDiamonds

You can prove the pythagorean theorem using that trigonomic identity but you have to prove that identity first, without using the pythagroean theorem. It would be hard since a lot of trigonometry is based on the pythag theorem

**Posted:** Wed May 13, 2009 2:43 pm