Question

1. triangle abc is congruent to triangles bad and cea.

a. explain why points d, a, and e are collinear.
b. explain why line de is parallel to line bc.

Explanation:

Step1: Use congruent triangle properties for angles

Since \(\triangle ABC\cong\triangle BAD\), then \(\angle BAD = \angle ABC\). Since \(\triangle ABC\cong\triangle CEA\), then \(\angle CAE=\angle BCA\). In \(\triangle ABC\), \(\angle ABC+\angle BAC+\angle BCA = 180^{\circ}\). Substituting \(\angle BAD\) for \(\angle ABC\) and \(\angle CAE\) for \(\angle BCA\), we get \(\angle BAD+\angle BAC+\angle CAE=\angle DAE = 180^{\circ}\).

Step2: Use congruent triangle properties for parallel lines

Since \(\triangle ABC\cong\triangle BAD\), \(AD = BC\). Since \(\triangle ABC\cong\triangle CEA\), \(AE = BC\), so \(AD = AE\). Also, from congruence \(\angle ADB=\angle BAC\) and \(\angle AEC=\angle BAC\). Then, using the converse of the alternate - interior angles theorem. \(\angle ADB=\angle DBC\) (from congruence \(\triangle ABC\cong\triangle BAD\)) and \(\angle AEC=\angle ECB\) (from congruence \(\triangle ABC\cong\triangle CEA\)). Also, since \(AD = AE\) and \(AD\parallel BC\), \(AE\parallel BC\) (because of angle - side - angle congruence implications). So, \(DE\) (composed of \(AD\) and \(AE\)) is parallel to \(BC\) as \(AD\parallel BC\) and \(AE\parallel BC\) and \(AD\) and \(AE\) are on the same line \(DE\).

Answer:

a. Because \(\angle DAE=\angle BAD + \angle BAC+\angle CAE\) and from congruent triangles \(\triangle ABC\cong\triangle BAD\) and \(\triangle ABC\cong\triangle CEA\), substituting angles gives \(\angle DAE = 180^{\circ}\), so \(D\), \(A\), \(E\) are collinear.
b. From congruent triangles \(\triangle ABC\cong\triangle BAD\) and \(\triangle ABC\cong\triangle CEA\), we have \(AD\parallel BC\) and \(AE\parallel BC\). Since \(D\), \(A\), \(E\) are collinear (from part a), \(DE\parallel BC\).