.please write it in a sum way not in words

Here is the solution for problem (6) in a step-by-step mathematical format:

Problem (6): Triangle ADE with parallel lines

Step 1: Interpret the diagram and state assumptions.

• The tick marks on $AD$ indicate $AB = BC = CD$.
• The tick marks on $AE$ indicate $AF = k$ (one tick) and $FG = GE = m$ (two ticks).
• For the problem to be solvable with the given angle relationships, we must assume that the lines $BF$, $CG$, and $DE$ are parallel. This implies that the segments on the transversals are proportional, which means $AB=BC=CD$ and $AF=FG=GE$.
• Therefore, we assume $AB=BC=CD=AF=FG=GE$. Let this common length be $L$.
• Given angles: $\angle AFG = 5y$, $\angle ADE = 2y$, $\angle AED = x$.

Step 2: Apply properties of parallel lines and similar triangles. If $BF \parallel CG \parallel DE$, then $\triangle ABF \sim \triangle ACG \sim \triangle ADE$. From similarity, corresponding angles are equal:

• $\angle ABF = \angle ACG = \angle ADE$
• $\angle AFB = \angle AGC = \angle AED$

Given $\angle ADE = 2y$, we have: $\angle ABF = \angle ACG = 2y$ Given $\angle AED = x$, we have: $\angle AFB = \angle AGC = x$

Step 3: Use angle relationships within $\triangle AFG$. Since we assumed $AF=FG=L$, $\triangle AFG$ is an isosceles triangle with base $AG$. The angles in $\triangle AFG$ are $\angle FAG$ (which is $\angle A$), $\angle AFG = 5y$, and $\angle FGA$. We know $\angle FGA = \angle AGC = x$. The sum of angles in $\triangle AFG$ is $180^\circ$: $\angle A + \angle AFG + \angle FGA = 180^\circ$ $\angle A + 5y + x = 180^\circ \quad (1)$

Step 4: Use angle relationships within $\triangle ADE$. The angles in $\triangle ADE$ are $\angle A$, $\angle ADE = 2y$, and $\angle AED = x$. The sum of angles in $\triangle ADE$ is $180^\circ$: $\angle A + \angle ADE + \angle AED = 180^\circ$ $\angle A + 2y + x = 180^\circ \quad (2)$

Step 5: Solve the system of equations. Subtract equation (2) from equation (1): $( \angle A + 5y + x ) - ( \angle A + 2y + x ) = 180^\circ - 180^\circ$ $3y = 0$ $y = \mathbf{0}$

Step 6: Conclusion. A value of $y=0$ is impossible for an angle in a triangle. This indicates that the given angle labels in the diagram are inconsistent with the geometric properties of parallel lines and similar triangles, even under the assumption that $BF \parallel CG \parallel DE$. Therefore, a numerical solution for $x$ and $y$ cannot be determined from the given information.