10. Let $A B C D$ be a parallelogram such that the bisector from $B$ intersects the side $C D$ at its midpoint $M$. The side $B C$ is 6 units long, and the diagonal $A C$ is 14 units long. Determine the length of $A M$.
(A) $2 \sqrt{19}$
(B) $14-3 \sqrt{3}$
(C) $4 \sqrt{5}$
(D) 9
(E) $2 \sqrt{22}$
10. The answer is $\mathbf{( C )}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c5c03d45c82457ddb3cag-14.jpg?height=540&width=1191&top_left_y=2180&top_left_x=478)

We observe that triangle $M B C$ is isosceles with base $M B$: indeed, $\widehat{C B M}=\widehat{M B A}$, by hypothesis, and $\widehat{A B M}=\widehat{B M C}$, as they are alternate interior angles with respect to the parallel lines $D C / / A B$. Therefore,

$$
\overline{A B}=\overline{D C}=2 \overline{M C}=2 \overline{C B}=2 \cdot 6=12
$$

Let $K$ be the projection of $C$ onto $A B$ and set $h=\overline{C K}$ and $d=\overline{B K}$. We find $h$ and $d$ through the system

$$
\left\{\begin{array}{l}
\overline{B C}^{2}-h^{2}=d^{2} \\
\overline{A C}^{2}-h^{2}=(\overline{A B}+d)^{2}
\end{array}\right.
$$

By subtracting the equations member by member, we obtain $d=2 / 3$ and $h=8 \sqrt{5} / 3$.

Let $H$ be the projection of $M$ onto $A B$, then

$$
\overline{A H}=\overline{A B}+\overline{B K}-\overline{H K},
$$

where $\overline{H K}=\overline{M C}$ (since $H K C M$ is a rectangle by construction) and $\overline{A B}=2 \overline{M C}$. It follows that $\overline{A H}=\overline{M C}+d=20 / 3$ and $\overline{A M}^{2}=(\overline{A H})^{2}+h^{2}=80$. Consequently, $\overline{A M}=4 \sqrt{5}$.

Second solution. As before, we have $\overline{A B}=2 \overline{B C}$. We observe that triangle $A B M$ is a right triangle at $M$: indeed, if $N$ is the midpoint of $\overline{A B}$, then $\overline{N M} / / \overline{B C}$, so $\overline{N M}=$ $\overline{A N}=\overline{N B}=6$, therefore $M$ lies on a circle of radius 6 centered at $N$ and in particular the angle at $M$ is right. Let $\beta$ be the angle $\widehat{A B M}$, then $\overline{A M}=\overline{A B} \sin \beta=12 \sin \beta$. Knowing the sides of triangle $A B C$, we can find $\cos (2 \beta)$ using the cosine rule: $\overline{A B}^{2}+\overline{B C}^{2}-$ $2 \overline{A B} \cdot \overline{B C} \cos (2 \beta)=\overline{A C}^{2}$, from which $\cos (2 \beta)=-1 / 9$. Using the cosine addition formula, we can find $\sin \beta$ from the equation $\cos (2 \beta)=1-2 \sin ^{2} \beta$, from which $\sin \beta=\sqrt{5} / 3$. Thus, $\overline{A M}=12 \cdot \sqrt{5} / 3=4 \sqrt{5}$.

Third solution. Using Heron's formula, we find that the area of triangle $A B C$ is $16 \sqrt{5}$. This area is also equal to the area of $A B M$, which in turn is twice the area of $A N M$. Let $2 x$ be the length of $\overline{A M}$, then we must have $1 / 2 \cdot 2 x \cdot \sqrt{6^{2}-x^{2}}=1 / 2 \cdot 16 \sqrt{5}=8 \sqrt{5}$. Squaring both sides and solving for $x^{2}$, we find $2 x=4 \sqrt{5}$.