Inscribing rhombus in a triangle's angle in only eight compass-and-straightedge steps
Draw circle $c$ of center $C$ and radius $CA$, meeting side $CB$ at $D$ (I'm assuming $AC<BC$).
Draw circle $b$ of center $B$ and radius $BD$, meeting side $AB$ at $E$.
Draw line $ED$: its second intersection with circle $c$ is point $X$ such that $CX=CA$ (obvious) and $CX\parallel AB$ (because triangles $CDX$ and $BDE$ have the same angles).
Draw line $AX$, meeting side $BC$ at $X'$, which is a vertex of the rhombus.
Draw circle $d$ of center $X'$ and radius $X'D$, meeting line $ED$ at $F$; line $X'F$ is parallel to $AB$, because triangles $X'DF$ and $BED$ have the same angles.
Draw line $AD$, meeting circle $d$ at $G$; line $X'G$ is parallel to $AC$, because triangles $X'DG$ and $CDA$ have the same angles.
- Draw lines $X'F$ and $X'G$ to complete the rhombus.
