I succeeded with analytical geometry,but not synthetical...Who can solve synthetical,pls?

Dear Mathlinkers, the consideration of the Haruki's lemme may help... Sincerely Jean-Louis

Any ideas? Sincerely Jean-Louis

Haruki's lemma is a good idea,thanks You,jayme! The circle of diameter AB intersect DP and CP in Q and R,respectively..The circumcicle of triangle PMR intersect AB ar S. Haruki's lemma leader to: AM*BN/MN=SB.....No,remain to prove that:SB=MN...Who can help me for this? If you desire,i can posted the analytical solution...

Dear soryn... the way Haruki is very interesting... SB = MN is the kern of this problem... Sincerely Jean-Louis

Dear jayme, your can help me?

Dear soryn, there are two equal triangles with bases SB and MN...I have to write my proof but I think there is more simple... Sincerely Jean-Louis

Just a question : which is the official solution? Sincerely Jean-Louis

soryn wrote: I succeeded with analytical geometry,but not synthetical...Who can solve synthetical,pls?

Hey sir here is the hint1Click to reveal hidden text

hint2 Click to reveal hidden text

hint 3 Click to reveal hidden text

Dear Mathlinker, can you precise the Hint 3... Very sincerely Jean-Louis

Here is the sol. Let $B',C'$ be the images of $B,C$ under the homothety sending $CD$ to $NM$. So $B'C'MN$ is a square. Note that $\Delta BNB' \sim \Delta C'MA$. So, $\frac{BN}{NB'}=\frac{C'M}{MA} \implies BN \cdot MA=NB' \cdot C'M=MN^2$

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Unexpectedly easy!Thank you,mr GeoKing! Here is my solution.Let be $A\left(0,0\right)$ the origin;then:$B\left(a,0\right);C\left(a,a\right);D\left(0,a\right)$ Let be $P\left(\alpha,\beta\right)$;then the vequation of line PD is :$y=\frac{x}{\alpha}\left(\beta-a\right)+a$ $y=0\Rightarrow x_{M}=\frac{\alpha a}{a-\beta}\Rightarrow\mid AM\mid=\frac{\alpha a}{a-\beta}$ After calculation,thevequation of vline CP is:$ y=x\cdotp\frac{\beta-a}{\alpha-a}+a-\frac{a\left(\beta-a\right)}{\alpha-a}$ $y=0\Rightarrow x_{N}=\frac{a\left(\alpha-\beta\right)}{a-\beta}\Rightarrow\mid BN\mid=a-x_{N}=\frac{a\left(\alpha-a\right)}{a-\beta}\Rightarrow\mid AM\mid\cdotp\mid BN\mid=\frac{a^{2}\cdotp\alpha\left(a-\alpha\right)}{\left(a-\beta\right)^{2}}$ $\mid MN\mid=\frac{a\left(\alpha-\beta\right)}{a-\beta}-\frac{\alpha a}{a-\beta}=-\frac{a\beta}{a-\beta}\Rightarrow\mid MN\mid^{2}=\frac{a^{2}\beta^{2}}{\left(a-\beta\right)^{2}}$ The equation of circle of diameter AB is:$\left(x-\frac{a}{2}\right)^{2}+y^{2}=\frac{a^{2}}{4}\Rightarrow\beta^{2}=\alpha\left(a-\alpha\right)$,done!

Thank you... Sincerely Jean-Louis