# Thread: Prove the following equation

1. ## Prove the following equation

Hi. How to prove that equation:

$\displaystyle |AB|^{2} = |QA| * |AE| + |NB| * |BD|$

$\displaystyle |BN| = |BC|$
$\displaystyle |AQ| = |AC|$

$\displaystyle |AD|$ and $\displaystyle |BD|$ are heights of acute-angled triangle $\displaystyle ABC$.

Illustration: http://i.imgur.com/9O8Bm.jpg

I would really appreciate any help.

2. ## Re: Prove the following equation

First point is that most of the diagram is unnecessary.
If $\displaystyle AQ=AC$ you don't need $\displaystyle Q$ or $\displaystyle P.$
If $\displaystyle BN=BC$ you don't need $\displaystyle N$ or $\displaystyle M.$
Further, $\displaystyle L$ and $\displaystyle K$ serve no purpose.
All that you need is the triangle $\displaystyle ABC$ and the perpendiculars $\displaystyle AD$ and $\displaystyle BE.$

Using Pythagoras twice, with the triangles $\displaystyle ADB$ and $\displaystyle ADC,$

$\displaystyle AB^{2}=AD^{2}+BD^{2}=(AC^{2}-CD^{2})+BD^{2}=AC^{2}-(BC-BD)^{2}+BD^{2}=AC^{2}-(BC^{2}+BD^{2}-2BC.BD)+BD^{2}=AC^{2}-BC^{2}+2BC.BD.$

Now repeat that sequence using the triangles $\displaystyle ABE$ and $\displaystyle BEC.$

Having done that, add the resulting equation to the one that's just been derived.

3. ## Re: Prove the following equation

Thank you very much. I've finally done that

Here are full calculations:

$\displaystyle AB^{2} = AD^{2} + BD^{2} = (AC^{2} - CD^{2}) + BD^{2} = AC^{2} - (BC - BD)^{2} + BD^{2} = AC^{2} - (BC^{2} + BD^{2} - 2 * BC * BD) + BD^{2} = AC^{2} - BC^{2} + 2 * BC * BD$

$\displaystyle AB^{2} = BE^{2} + AE^{2} = (BC^{2} - CE^{2}) + AE^{2} = BC^{2} - (AC - AE)^{2} + AE^{2} = BC^{2} - (AC^{2} + AE^{2} - 2 * AC * AE) + AE^{2} = BC^{2} - AC^{2} + 2 * AC * AE$

$\displaystyle 2AB^{2} = AC^{2} - BC^{2} + 2 * BC * BD + BC^{2} - AC^{2} + 2 * AC * AE$
$\displaystyle 2AB^{2} = 2 * BC * BD + 2 * AC * AE$
$\displaystyle AB^{2} = BC * BD + AC * AE$