Help! - Pythagorean Theorem Proof

**ReneePatt** · December 11th 2009, 06:40 AM

In need of help. I've searched the internet and can't seem to find a pythag proof that makes since to me.

Looking at the diagram below and using the following theorem, how do you prove the pythagorean theorem.

Theorem: The power of a point is well defined; that is, the same value is obtained regardless of which line l is used in the definition as long as the line has at least one point of intersection with the circle.

I'd even settle for just proving the pythagorean theorem without using the above theorem.

Prove: (AD)(DC) = (AE)(EB) + (BF)(FC)

Help! - Pythagorean Theorem Proof-pythag-theorem.jpg

**mathaddict** · December 11th 2009, 07:00 AM

Originally Posted by ReneePatt

In need of help. I've searched the internet and can't seem to find a pythag proof that makes since to me.

Looking at the diagram below and using the following theorem, how do you prove the pythagorean theorem.

Theorem: The power of a point is well defined; that is, the same value is obtained regardless of which line l is used in the definition as long as the line has at least one point of intersection with the circle.

I'd even settle for just proving the pythagorean theorem without using the above theorem.

Prove: (AD)(DC) = (AE)(EB) + (BF)(FC)

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HI

Take a look at the diagram i attached . $\triangle ABC$ with right angle at B . Then construct BD which is perpendicular to AC . We are trying to prove that $AC^2=AB^2+BC^2$

so here $\triangle ADB$ , $\triangle ABC$ , $\triangle BDC$ are similar .

$\frac{AD}{AB}=\frac{AB}{AC}\Rightarrow AB^2=AD\cdot AC$

$\frac{BD}{DC}=\frac{AC}{BC}\Rightarrow BC^2=AC\cdot DC$

$\therefore AB^2+BC^2=AD\cdot AC+AC\cdot DC=AC(AD+DC)=AC\cdot AC$

hence proved .

**ReneePatt** · December 11th 2009, 07:08 AM

Originally Posted by mathaddict

so here $\triangle ADB$ , $\triangle ABC$ , $\triangle BDC$ are similar .

Hi,
You say that $\triangle ADB$ , $\triangle ABC$ , $\triangle BDC$ are similar but there isn't a $\triangle BDC$ . Did you mean $\triangle ADC$ ?

**mathaddict** · December 11th 2009, 07:16 AM

Originally Posted by ReneePatt

Hi,
You say that $\triangle ADB$ , $\triangle ABC$ , $\triangle BDC$ are similar but there isn't a $\triangle BDC$ . Did you mean $\triangle ADC$ ?

hey the diagram i attached previously is incorrect . This is the correct one . I accidentally swapped the A and B .

**ReneePatt** · December 11th 2009, 07:39 AM

Originally Posted by mathaddict

hey the diagram i attached previously is incorrect . This is the correct one . I accidentally swapped the A and B .

Thank you very much!!

This makes so much sense.

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