Question

1. given: ( overline{ac} ) bisects ( angle dab ); ( overline{ac} ) bisects ( angle dcb ) prove: ( overline{bc} cong overline{dc} ) ( overline{ac} ) bisects ( angle dab ) ( overline{ac} ) bisects ( angle dcb ) given

Explanation:

Step1: Use the angle - bisector property

Since \(\overline{AC}\) bisects \(\angle DAB\), we have \(\angle1=\angle2\). Since \(\overline{AC}\) bisects \(\angle DCB\), we have \(\angle3 = \angle4\). Also, \(AC=AC\) (common side).

Step2: Apply the ASA (Angle - Side - Angle) congruence criterion

In \(\triangle ADC\) and \(\triangle ABC\), \(\angle1=\angle2\) (from angle - bisector), \(AC = AC\) (common side), \(\angle3=\angle4\) (from angle - bisector). By ASA congruence, \(\triangle ADC\cong\triangle ABC\).

Step3: Use the property of congruent triangles

If \(\triangle ADC\cong\triangle ABC\), then the corresponding sides are equal. So, \(BC = DC\) (corresponding parts of congruent triangles).

Answer:

\(BC = DC\) is proved as \(\triangle ADC\cong\triangle ABC\) by ASA congruence criterion and then using the property of congruent triangles.