In the given figure, if $AD=(x+3)$ cm, $DB=(3x+19)$ cm, $AE=x$ cm, $EC=(3x+4)$ cm and $DE \parallel BC$, then the value of $x$ is:
Show Hint
When a line through two sides of a triangle is parallel to the third side, use similarity (or the basic proportionality theorem) to relate parts to wholes.
Step 1: Use similarity (Basic proportionality).
Since $DE \parallel BC$, triangles $ADE$ and $ABC$ are similar. Hence, the corresponding side ratios are equal:
\[
\frac{AD}{AB}=\frac{AE}{AC}.
\]
Step 2: Express the whole sides.
Along $\overline{AB}$: $AB = AD + DB = (x+3) + (3x+19) = 4x + 22$.
Along $\overline{AC}$: $AC = AE + EC = x + (3x+4) = 4x + 4$.
Step 3: Set up and solve the proportion.
\[
\frac{x+3}{4x+22}=\frac{x}{4x+4}
\;\Rightarrow\;
(x+3)(4x+4)=x(4x+22).
\]
\[
4(x+1)(x+3)=4x^2+22x
\;\Rightarrow\;
4x^2+16x+12=4x^2+22x
\;\Rightarrow\;
12=6x
\;\Rightarrow\;
x=2.
\]
Step 4: Conclusion.
Thus, $x=2$ cm.
