We have pyramid MABC, where BC=x, AM=y, and AB=AC=MB=MC=z. MK is height in triangle CMB, and AK is height in triangle ABC. They will both cross the point K, because they share base CB and they are both isosceles triangle, legs being equal to z and for the same reason AK=MK. MO is height in a triangle AKM, but how do we know that MO is also the height of the pyramid?
Notice that $BC$ is orthogonal to plane $AMK$ because it's orthogonal to two lines in $AMK$: $MK$ and $AK$. Let's draw a line in plane $ABC$ thru point $O$ parallel to $BC$, it will also be perpendicular to plane $AMK$ and perpendicular to $MO$. Now we have that $MO$ is perpendicular to two intersecting lines in plane $ABC$ (the other one is $AK$) and thus it's perpendicular to the plane. Done.