Problem 9.3. Given a convex quadrilateral $ABCD$, $X$ is the midpoint of diagonal $AC$. It turns out that $CD \parallel BX$. Find $AD$, if it is known that $BX=3, BC=7, CD=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2befe970655743580344g-02.jpg?height=224&width=495&top_left_y=1341&top_left_x=479)
Answer: 14.

Solution. Double the median $B X$ of triangle $A B C$, to get point $M$. Quadrilateral $A B C M$ is a parallelogram (Fig. 1).

Notice that $B C D M$ is also a parallelogram, since segments $B M$ and $C D$ are equal in length (both 6) and parallel. This means that point $M$ lies on segment $A D$, since $A M \| B C$ and $M D \| B C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2befe970655743580344g-03.jpg?height=262&width=601&top_left_y=91&top_left_x=420)

Fig. 1: to the solution of problem 9.3

Now it is easy to find the desired segment:

$$
A D=A M+M D=B C+B C=14
$$