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Name the large triangle as △ABC, ∠A is the top angle, ∠B on the left and ∠C on the right.
Name E is the joint point on AB, D is the joint point on AC, F is the joint of BD and CE.

1)
Considering △BEF and △DEF share the same height if BD is the base, u/y=|BF|/|DF|.
Considering △BCF and △DCF share the same height if BD is the base, z/v=|BF|/|DF|.
Therefore, u/y=z/v, i.e. uv=yz.

2)
Considering △ADE and △BDE share the same height if AB is the base, $x/(y+u)=|AE|/|BE|.$
Considering △ACE and △BCE share the same height if AB is the base, $(x+y+v)/(u+z)=|AE|/|BE|.$
Therefore, $x/(y+u)=(x+y+v)/(u+z)$ or equivalently
$xu+xz=y(x+y+v)+ux+uy+uv$
We have already $uv=yz$, so
\begin{align*}
xz&=y(x+y+v)+yu+yz\\
&=y(x+y+z+u+v)
\end{align*}
or equivalently
$x+y+z+u+v=\frac{xz}{y}$