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$ABCD$ is a parallelogram. If $AB=6,AC=7,DE=2$.$CF= a/b$ and $gcd(a,b)=1$. then, $a+b$=?

Let, Area of $\triangle ABC$ be denoted by $(ABC)$
$(ABC) = (ADC)$
$AC \times DE = AB \times CF$
The rest is up to you.

Answer will be 10

This problem appeared as a problem in Regional contest (Junior Category) 2020. Okay, Here is the full solution:
Think about the triangle $ADC$, in triangle $ADC$, height of the triangle is $DE=2$ and the land is $AC=7$. Then the area of triangle $ADC$ is: $\frac{2*7}{2} = 7$. Now, we know that, the diagonal of a parallelogram divides the entire parallelogram into half. It means the total area of parallelogram $ABCD$ is 2 times the area of triangle $ADC = 7*2 = 14$. Now, the next step is simple. Lets think about the entire parallelogram $ABCD$. We determined that the area of parallelogram $ABCD$ is 14. The height of parallelogram $ABCD$ is $CF$ and land is $AB$. Now, it means,
$CF * AB = 14$
$ CF*6=14$
$ CF = \frac{14}{6}$
$CF = \frac{7}{3}$
Then, CF = $\frac{7}{3} CF=\frac{a}{b}$
gcd(7,3)= 1. Then a=7, b=3. $a+B=7+3=10 $
So, the answer is 10
By the way, it is also my first post