Help. ​

Tangents drawn to a circle from the same point outside the circle are equal.....therefore....

JA = JB, LA = LC, KC = KB  ..so

JA + JB + LA + LC + KC + KB  = perimeter of triangle JKL....so...by substitution

JA + JA + LA + LA + KC + KC   =  perimeter of triangle JKL

2 ( JA + LA + KC)  =  perimeter of triangle JKL

2 ( 13 + 19 + 7)  =  perimeter of triangle JKL

2 ( 39)  =  perimeter of triangle JKL

78  =  perimeter of triangle JKL