Problem 6. A convex quadrilateral $ABCD$ is such that $\angle BAC = \angle BDA$ and $\angle BAD = \angle ADC = 60^{\circ}$. Find the length of $AD$, given that $AB = 14$ and $CD = 6$.

Answer: 20.
Solution. Extend $A B$ and $C D$ to intersect at point $P$. Since $\angle P A D = \angle A D P = 60^{\circ}$, triangle $A D P$ is equilateral. Next, we note that triangle $A P C$ is congruent to triangle $D A B$, because $A P = A B$, $\angle A P C = 60^{\circ} = \angle D A B$, and $\angle P A C = \angle A D B$ (Fig. 2). Therefore, $P C = A B = 14$, and $A D = P D = P C + C D = 14 + 6 = 20$.

![](https://cdn.mathpix.com/cropped/2024_05_06_23b404dba1f5d039cdf1g-4.jpg?height=450&width=495&top_left_y=251&top_left_x=479)

Fig. 2: to the solution of problem 6

## Criteria

0 points. Only the correct answer without justification.

1 point. The additional construction to the equilateral triangle $A P D$ is present, but there is no further progress.

3 points. A pair of congruent triangles leading to the solution is found, but the solution is not completed.

4 points. The correct answer and justification are provided.