Can someone help me? I do not know how to continue with the problem.

Question

Parallelogram ABCD is given. The angle bisector of $\angle DAB$ intersects the side $\overline{DC}$ in the point L and the diagonal in the point K. Let $\overline{DK}:\overline{KB}=3:4$. Calculate the length of $\overline{LC}$ if the perimeter od the parallelogram is 28.

From the perimeter i found the sum $a+b=14$ (a,b are the sides of parallelogram where 'a' is the longer side). I marked $\overline{DK}:\overline{KB}$ like $\overline{DK}:\overline{KB}=3k:4k$ so the diagonal DB is: $\overline{DB}=3k+4k=7k$. What should i do now?

Answer 1

Let $AB=a$ and $CL=x$.

Thus, $LD=a-x$, $AD=14-a$ and since $\Delta ABK\sim\Delta LDK$, we obtain: $$\frac{a}{a-x}=\frac{4}{3}.$$ Also, we have $$\frac{a}{14-a}=\frac{4}{3},$$ which gives $a=8$ and $x=2$.

Answer 2

Hint:

• Due to Angle bisector theorem we have that $|DA| : |AB| = 3 : 4$.
• Because $AB$ and $DC$ are parallel, $|\angle DLA| = |\angle LAB| = |\angle DAL|$.

I hope this helps $\ddot\smile$