1) Looking at that circle, we have two right triangles within. And Since SN is perpendicular to JK we can state that SN bisects JK. Therefore we have a
Δ JNS
2) By the same principle, we have SP bisecting, the Line segment LM into two equal segments therefore PL = 6
3) As we have 4 similar triangles SAS: Δ SPL, ΔSPM, ΔSNJ, ΔSNK we can write this equation:
JK≅LM
2x-2=12
2x=12+2
2x =14
x=7
