There is a formula (sometimes called Apollonius's theorem or a special case of Stewart's theorem) that gives the square of a median in terms of the three side lengths. Can you write $AD^2$ in terms of $AB$, $AC$, and $BC$?

Let the side that the median lands on be $a$, and the other two sides be $b$ and $c$. Then substitute $a=BC$, $b=AC$, and $c=AB$ into the median formula and simplify step by step.

Once you find $AD^2$ as a number, take its square root and factor it to see it as $2\sqrt{\text{(something)}}$. Then match that exact form to one of the choices.

In a Desmos line, type the expression for $AD^2$ using the median formula:

(2*15^2 + 2*13^2 - 14^2)/4

Desmos will display the numerical value of $AD^2$.

In a new line, type sqrt((2*15^2 + 2*13^2 - 14^2)/4) to get the decimal value of $AD$.

To see which option matches this length, type each choice into Desmos (for example, 2*sqrt(33), 2*sqrt(35), etc.) and compare their decimal values to the value from the previous step. The choice whose value exactly matches is the correct answer.

Point $D$ is the midpoint of $\overline{BC}$, so $BD=DC=\dfrac{BC}{2}=7$. This means $AD$ is a median from vertex $A$ to side $BC$.

For a triangle with side lengths $a$, $b$, and $c$, and median from the vertex between sides $b$ and $c$ to side $a$, the median length $m_a$ satisfies:

Here, $a$ is the side the median hits (opposite the vertex), and $b$ and $c$ are the other two sides.

In $\triangle ABC$:

• The median is drawn from $A$ to side $BC$, so let $a = BC = 14$.
• The other sides are $b = AC = 15$ and $c = AB = 13$.

Substitute these into the median formula:

Now simplify step by step:

• $15^2 = 225$, so $2(15^2) = 2 \cdot 225 = 450$
• $13^2 = 169$, so $2(13^2) = 2 \cdot 169 = 338$
• $14^2 = 196$

Plug these in:

So the median satisfies $AD^2 = 148$.

To find $AD$, take the square root of 148:

Factor 148:

• $148 = 4 \cdot 37$
• $\sqrt{148} = \sqrt{4 \cdot 37} = 2\sqrt{37}$

So the length of $\overline{AD}$ is $2\sqrt{37}$, which matches answer choice C.