Given:
AB || DE
BC = 14
CD = 7
CE = 7
DE = 5
To prove:
The two triangles are congruent.
Consider the two triangles ABC and EDC.
[tex]\begin{gathered} \angle BCA=\angle\text{DCE ( Vertical angles)} \\ \angle ABC=\angle EDC\text{ (Alternate angles)} \\ \angle BAC=\angle DEC\text{ (Alternate angles)} \end{gathered}[/tex]
By AAA similarity, If any two angles of a triangle are equal to any two angles of another triangle, then the two triangles are similar to each other.
Therefore,
[tex]\Delta ABC\approx\Delta EDC[/tex]
So, the corresponding sides are proportional.
[tex]\begin{gathered} \frac{AB}{ED}=\frac{BC}{DC}=\frac{AC}{EC} \\ \frac{AB}{5}=\frac{14}{2}=\frac{AC}{7} \\ \frac{AB}{5}=7=\frac{AC}{7} \end{gathered}[/tex]
So, equating (1) and (2), we get
AB = 35
So, equating (2) and (3), we get
AC = 49
Thus, the missing sides are AB=35 and AC=49.
