A Quick Triangle Problem

By the triangle inequality, $AC+AD>CD$ , so $5+AD>15$ or $AD>10$ . But, $AB+BD>AD$ , so $4+8=12>AD$ . Therefore, $10<AD<12$ , and the only integer satisfying this is $AD=\boxed{11}$ .

sum of the two sides of a triangle must be greater than the third side:

in this case- AD<(8+4)<(5+15) AD<12<20 so AD<12

now, [as the sides are integer] when AD belongs to (1-10) : triangle ABD and triangle ACD cannot be formed.

when AD = 12
then triangle ABD cannot be formed.

so we are left with only one choice AD = 11

In Triangle ABD, AD less than 12, But in Triangle ADC, AD is greater than 10, So 11 is the answer.

Triangle inequality: 8 - 4 < AD < 8 + 4 15 - 5 < AD < 15 + 5 4 < AD < 12 10 < AD < 20 Only AD = 11 works.

From $\triangle ABD \text{, } 4 < AD < 12 \text{, but from } \triangle ADC \text{, } 10 < AD < 20 \text{. Thus, } AD = \boxed{11} \text{.}$