Equilateral triangle $A B C$ has sides of length 4. The midpoint of $B C$ is $D$, and the midpoint of $A D$ is $E$. The value of $E C^{2}$ is
(A) 7
(B) 6
(C) 6.25
(D) 8
(E) 10

## Part C: Each correct answer is worth 8.
Since $\triangle A B C$ is equilateral and has sides of length 4 , then $A B=B C=A C=4$.

The midpoint of $B C$ is $D$ and so $B D=C D=2$.

The midpoint of $A D$ is $E$ and so $A E=E D$.

Since $A B=A C$ and $D$ is the midpoint of $B C$, then $A D$ is

perpendicular to $B C$, as shown.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-037.jpg?height=285&width=296&top_left_y=1885&top_left_x=1554)

Triangle $A D C$ is a right-angled triangle, and so by the Pythagorean Theorem, we get $(A C)^{2}=(A D)^{2}+(D C)^{2}$ or $4^{2}=(A D)^{2}+2^{2}$, and so $(A D)^{2}=16-4=12$.

Similarly, $\triangle E D C$ is right-angled, and so by the Pythagorean Theorem, we get $(E C)^{2}=(E D)^{2}+(D C)^{2}$ or $(E C)^{2}=(E D)^{2}+2^{2}$.

Since $E D=\frac{1}{2} A D$, then $(E D)^{2}=\frac{1}{2} A D \times \frac{1}{2} A D$ or $(E D)^{2}=\frac{1}{4}(A D)^{2}$.

Since $A D^{2}=12$, then $(E D)^{2}=\frac{1}{4} \times 12=3$.

Substituting, we get $(E C)^{2}=3+2^{2}$, and so $(E C)^{2}=7$.