This is my exercise:
We are given a paper in the form of a triangle, with two sides being 6 and 7 units long. We first fold the triangle so that the corner $B$ is on the side $AC$. We then unfold it so that the corner $A$ is on the side $BC$. If we then unfold the paper, the creases are 4 and 2 units long (as shown in the picture). Determine $x$.
Now, I am somewhat confused by the question. How can we calculate the part $x$ if we aren't explicitly told that the creases form $90$ degree angles somewhere in the triangle?
From the first fold, let $AD$ be the crease from $A$, where $D$ is on side $BC$. From the second fold, let E be the crease from $B$, where $E$ is on side $AC$. Let $I$ be the intersection of $AD$ and $BE$.
Note that the creases $AD$ and $BE$ bisect corners $A$ and $B$ respectively.
Apply the angle bisector theorem on $\triangle ABD$ with angle bisector $BI$, then
$$\begin{align*} \frac{|BD|}{|BA|} &= \frac{|DI|}{|AI|}\\ |BD| &= \frac{2}{4}\cdot 6 = 3\\ |DC| &= 7 - |BD| = 4 \end{align*}$$
Apply the angle bisector theorem on $\triangle ABC$ with angle bisector $AD$, then
$$\begin{align*} \frac{|AC|}{|AB|} &= \frac{|CD|}{|BD|}\\ |AC| &= \frac{4}{3}\cdot 6 = 8 \end{align*}$$
Apply the angle bisector theorem on $\triangle ABC$ with angle bisector $BE$, then
$$\begin{align*} \frac{|AE|}{|CE|} &= \frac{|BA|}{|BC|} = \frac67\\ |AE|+|CE| &= |AC| = 8\\ x &= |AE|\\ &= \frac{6}{6+7}\cdot 8\\ &= \frac{48}{13} \end{align*}$$