A shade crossed

This triangle is a right angled isosceles triangle, the hypotenuse being $\sqrt{2}b$. We draw a line through $A$ and point $M$, the midpoint of the line $BC$. We draw the line $B`C`$ giving another right angles isosceles triangle $AB`C`$, similar to triangle $ABC$ but with sides a and hypotenuse $\sqrt{2}a$. Now $N$ is the point where the line $B`C`$ meets the line $AM$, and $P$ is the point where $BC`$ meets $B`C$ (also on $AM$).

It is clear that triangle $PBC$ is similar to triangle $P B`C`$ and the enlargement factor from $B`C`$ to $BC$ is $b/a$. So the line $PM$ is $b/a$ times as long as the line $PN$. Also the line $AN$ is half the length of $B`C`$, so it is $\sqrt{2}a/2$. The line $AM$ is half the length of $BC$ so it is $\sqrt{2}b/2$.