$M$ is the midpoint of the chord $AB$ of circle $C(O,r)$. Show that if a different chord $ CD $ contains $M$, then $ AB < CD $ .(You may use Pythagoras's Theorem)
Let $ \Delta CBM\sim \Delta ADM $, $ BM = x $ and $MA =x$
I have that $\frac{CM}{MA} = \frac{BM}{DM} \rightarrow x^2 = (CM)(MD) $
by$ A.M. - G.M.$ i have $ CM+MD \geq 2x $
if $ CM + MD = 2x $ then chord $CD $ is chord $AB$.
if $ CM+ MD > 2x $ then chord $CD$ is other chord that is not chord $AB$ .
so I can conclude that $ AB< CD $.
I want to know if there's a way to prove this using Pythagoras's theorem or other way, and, is my solution correct ?
Your proof is correct, here's another possible way.
$OM$ is perpendicular to $AB$ and is not perpendicular to $CD$. Hence the midpoint $N$ of $CD$ is different from $M$ and $ON<OM$ (because in right triangle $ONM$ side $OM$ is the hypothenuse). It follows then from Pythagoras's theorem: $$ CN=\sqrt{r^2-ON^2}>\sqrt{r^2-OM^2}=AM. $$