Question

question
given: (abcd) is a parallelogram and (\triangle abc cong \triangle dcb).
prove: (overline{ab} perp overline{bc}).

Explanation:

Step1: Recall properties of parallelogram

In parallelogram \(ABCD\), \(AB \parallel DC\) and \(AB = DC\), \(BC\) is a common side. Also, \(\angle ABC + \angle DCB = 180^\circ\) (consecutive angles in parallelogram are supplementary).

Step2: Use congruent triangles property

Since \(\triangle ABC \cong \triangle DCB\), corresponding angles \(\angle ABC=\angle DCB\).

Step3: Solve for \(\angle ABC\)

Let \(\angle ABC = x\) and \(\angle DCB = x\) (from congruence). Then \(x + x=180^\circ\) (from supplementary angles in parallelogram), so \(2x = 180^\circ\), which gives \(x = 90^\circ\).

Step4: Conclude perpendicularity

If \(\angle ABC = 90^\circ\), then \(\overline{AB}\perp\overline{BC}\) (definition of perpendicular lines: angle between them is \(90^\circ\)).

Answer:

By using properties of parallelogram (consecutive angles supplementary) and congruent triangles (\(\triangle ABC \cong \triangle DCB\) implies \(\angle ABC=\angle DCB\)), we find \(\angle ABC = 90^\circ\), hence \(\overline{AB}\perp\overline{BC}\).