Question

question 15
not necessarily drawn to scale:
given: ab // de and ab = ed
prove: \\( \triangle abc \cong \triangle edc \\)
proof:
ab // de given
ab = ed given
\\( \angle abc \cong \angle edc \\) alternate interior angles of parallel lines
\\( \angle acb \cong \angle ecd \\) select an answer
\\( \triangle acb \cong \triangle ecd \\) aas

Explanation:

Step1: Vertical angles

Vertical angles are formed when two lines intersect. In this case, lines \(AE\) and \(BD\) intersect at point \(C\). By the vertical - angles theorem, \(\angle ACB\) and \(\angle ECD\) are vertical angles.

Step2: Vertical - angles theorem

The vertical - angles theorem states that vertical angles are congruent. So, \(\angle ACB\cong\angle ECD\) because they are vertical angles.

Answer:

Vertical Angles