Chords DE and FG intersect at X, and so (DX)(EX)=(FX)(GX) or (DX)(8)=(6)(4).
Solving this equation for DX, we get DX=8(6)(4)=824=3.
The length of DX is 3.
Chords JK and LM intersect at X, and so (JX)(KX)=(LX)(MX) or(8y)(10)=(16)(y+9).
Solving this equation, we get 80y=16y+144 or 64y=144, and so y=64144=49.
Chords PQ and ST intersect at U, and so (PU)(QU)=(SU)(TU).
Since TU=TV+UV, then TU=6+n.
Substituting values into (PU)(QU)=(SU)(TU), we get (m)(5)=(3)(6+n), and so 5m=18+3n.
Chords PR and ST intersect at V, and so (PV)(RV)=(TV)(SV).
Since SV=SU+UV, then SV=3+n.
Substituting values into (PV)(RV)=(TV)(SV), we get (n)(8)=(6)(3+n), and so 8n=18+6n or 2n=18 or n=9.
Substituting n=9 into 5m=18+3n, we get 5m=18+3(9) or 5m=45, and so m=9.
Therefore, m=9 and n=9.