In a triangle $ABC$, $AB=5$, $BC=16$, $AC= \sqrt{153}$, and $D$ is on segment $BC$. Compute the sum of all possible integral measures of $AD$.
I've been having trouble trying to solve this problem and any help would be greatly appreciated. If you can provide a solution for me to study, thank you in advance.
Since $BC^2 = 256 > 178 = AB^2+AC^2$, triangle $ABC$ is obtuse with $\angle BAC > 90^{\circ}$. If a different angle was obtuse, the solution below would be slightly different.
The shortest possible value for $AD$ will be when $AD \perp BC$, i.e. $AD$ is the altitude to side $BC$. Use Heron's formula to find the area of the triangle. Then, this altitude satisfies $\dfrac{1}{2} \cdot AD \cdot BC = \text{Area}$.
The longest possible value for $AD$ will occur either when $D$ is an endpoint of $BC$. It should be obvious which endpoint yields the largest value.
Now, you just have to add up all the integers between the smallest and largest values of $AD$.
$AC = \sqrt{153} = 12.xxxxx$
∴ the largest integral value of AD is 12
By cosine law, $\cos \angle ABC = … = 0.8$
∴ $\sin ABC = … = 0.6$
The shortest value of AD is $AB \sin \angle ABC = ... = 3$ (why?)
Hence, $3 \le integral- value -of (AD) \le 12$
