given is a triangle with corners A,B,C and corresponding sides a, b, c.
D is a point somewhere between A and B.
I have to proof:
$b^2|BD| +a^2|AD| - c|CD|^2 = c|AD||BD|$
Unfortunately I have no idea - I just tried with some pythagoras but only got nonsense.
So I'll be glad if you can give me a hint where to start, then I might get it :)
Kind regards!
You need at the fondation of your work the cosinus rule (Al Kashi), which is the extension of Phytagoras.
That is (1) $$a^2=b^2+c^2-2bc \cos \angle A$$
The same way you have (2) $$|CD|^2=b^2+|AD|^2-2b|AD|\cos \angle A$$
Then
$b^2|BD| +a^2|AD| - c|CD|^2 =a^2|AD|+b^2(c-|AD|)-c(b^2+|AD|^2-2b|AD|\cos \angle A)$
$b^2|BD| +a^2|AD| - c|CD|^2 =(a^2-b^2)|AD|-c|AD|^2+2bc|AD|\cos \angle A$
$b^2|BD| +a^2|AD| - c|CD|^2 =(c^2-2bc\cos \angle A)|AD|-c|AD|^2+2bc|AD|\cos \angle A$
$b^2|BD| +a^2|AD| - c|CD|^2 =c|AD|(c-|AD|)=c|AD||BD|$