Which set of segment lenghts could be used to construct a triangle

The complete question in the attached figure

we know that
To construct a triangle, two segments added together have to be greater than the 3rd segment, 
so
case A) 
2, 12, 19 
2+12 < 19------> these segments can not be used to form a triangle

case B) 
4, 7, 13
4+7 < 13------> these segments can not be used to form a triangle

case C) 
4, 5, 12 
4+5 < 12------> these segments can not be used to form a triangle

case D) 
13, 2, 12
13+2 > 12
13+12> 2
2+12>13
so
these segments can be used to form a triangle

the answer is
the option D) 
13, 2, 12
these segments can be used to form a triangle

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12