sorry for the drawing. From a point $D$ on side $AB$, a line $DE$ is drawn through a point $E$ on side $AC$ such that angle $AED$ is equal to angle $ABC$. If the perimeter of the triangle $ADE$ is equal to 56. Find the sum of the length of line segments $BD$ and $CD$
I have my solution and used the perimeter formula. where $\frac{P1}{P2}$$=$$\frac{x1}{x2}$ so I used $\frac{56}{140}$$=$$\frac{AD}{30}$ so I got 12 and $\frac{56}{140}$$=$$\frac{AE}{50}$ so $BD$ = $AB-AD$ therefore $30-12$=$18$ It's wrong that I got $18$ but how do I get the correct value?
You assumed that angle $ABC$ was the same as angle $ADE$ which would then give you the correct answer by the method you were using. But in fact angle $ABC$ is equal to angle $AED$ not angle $ADE$. You still get a similar triangle, but you need to associate the correct sides.
So side $AD$ corresponds to side $AC$. Thus: $\frac{56}{140} = \frac{AD}{50} \Leftrightarrow AD = 20$. I'm sure you can do the rest.