Problem 7.7. On the side $AC$ of an equilateral triangle $ABC$, a point $D$ is marked. On the segments $AD$ and $DC$, equilateral triangles $ADE$ and $DCF$ are constructed outward from the original triangle. It is known that the perimeter of triangle $DEF$ is 19, and the perimeter of pentagon $ABCFE$ is 43.

![](https://cdn.mathpix.com/cropped/2024_05_06_4d36ea080cd7b65cb574g-6.jpg?height=504&width=389&top_left_y=1399&top_left_x=528)
(a) (2 points) Find the length of the segment $AB$.

(b) (2 points) Find the length of the segment $EF$.
# Answer:

(a) $(2$ points) $A B=12$.

(b) $(2$ points) $E F=7$.

Solution. Let $A E=E D=D A=a, D F=F C=C A=b$, then $A B=B C=C A=a+b$. According to the condition, $19=a+b+E F$ and $43=(a+b)+(a+b)+b+E F+a$. Subtract the first equation from the second and divide by two, we get $12=\frac{43-19}{2}=a+b=A B$. Then from the first equation, we find $E F=19-(a+b)=19-12=7$.