How can we show that at least one of the triangles $PAB,PBC,PAC$ has a smaller perimeter than $ABC$?

Question

For any triangle $ABC$ on the plane, we draw a circle $\Omega$ with center $O$ such that $AB$ is its diameter. We place a point $P$ inside $\Omega$. How can we show that at least one of the triangles $PAB,PBC,PAC$ has a smaller perimeter than $ABC$? I have no idea how to do this, can this be proved with simple geometry?

Answer 1

hint: consider a triangle $EAC$, see picture below:

we want to prove $EB+EC < AB+AC \iff EB+OC < OE+AC \iff EB-AC < OE-OC$

Flip $C$ to $C' $ $ \implies AC=BC' ,OC=OC' ,EB-AC < OE-OC \iff EB-BC'< OE-OC'=EC'$

now prove $PB+PC <EB+EC$, you can go on now. it is easy to let $P$ on the circle first. then inside circle.