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:

a.

Since \(\triangle ABC\cong\triangle BAD\) and \(\triangle ABC\cong\triangle CEA\), we know that \(\angle BAC=\angle ABD\) and \(\angle BAC = \angle ACE\). Also, \(\angle BAD=\angle ABC\) and \(\angle CAE=\angle ACB\).

By the angle - sum property of a triangle (\(\angle ABC+\angle BAC+\angle ACB = 180^{\circ}\)), and because \(\angle DAB=\angle ABC\), \(\angle BAC\) is common (in congruent triangles), and \(\angle EAC=\angle ACB\), we have \(\angle DAB+\angle BAC+\angle CAE=\angle ABC+\angle BAC+\angle ACB = 180^{\circ}\).

Since \(\triangle ABC\cong\triangle BAD\), then \(AD = BC\). Since \(\triangle ABC\cong\triangle CEA\), then \(AE=BC\). So \(AD = AE\).

Also, \(\angle ADB=\angle BCA\) (from \(\triangle ABC\cong\triangle BAD\)) and \(\angle AEC=\angle ABC\) (from \(\triangle ABC\cong\triangle CEA\)).

Since \(AD = AE\) (proven above) and \(\angle DAB+\angle BAC+\angle CAE = 180^{\circ}\) (from part a), and using the properties of congruent triangles \(\angle ADB+\angle ABD+\angle BAD=180^{\circ}\), \(\angle AEC+\angle ACE+\angle CAE = 180^{\circ}\), \(\angle ABC+\angle BAC+\angle ACB=180^{\circ}\)

We can use the alternate - interior angles or the slope concept (if we consider coordinate geometry, but using triangle congruence):

Since \(\triangle ABC\cong\triangle BAD\) and \(\triangle ABC\cong\triangle CEA\), \(\angle ADB=\angle BCA\) and \(\angle AEC=\angle ABC\). Also, \(AD = BC\) and \(AE = BC\) (so \(AD=AE\)).

By the converse of the alternate - interior angles theorem: If two lines are cut by a transversal and the alternate - interior angles are equal, then the lines are parallel.

Let's consider the transversal \(BD\) (for \(DE\) and \(BC\)): \(\angle ADB=\angle BCA\) (from \(\triangle ABC\cong\triangle BAD\)) and the transversal \(CE\): \(\angle AEC=\angle ABC\) (from \(\triangle ABC\cong\triangle CEA\)).

Another way: Since \(AD = BC\) and \(AE=BC\), \(AD = AE\). Also, \(\angle DAB+\angle BAC+\angle CAE = 180^{\circ}\)

Since \(\triangle ABC\cong\triangle BAD\), \(\angle ABD=\angle BAC\) and since \(\triangle ABC\cong\triangle CEA\), \(\angle ACE=\angle BAC\)

We know that \(DE=AD + AE=2BC\) (since \(AD = BC\) and \(AE = BC\))

By the mid - point theorem (a special case, since we can construct parallelograms in a sense, because of the equal side lengths and angle relations from congruent triangles)

Since \(\triangle ABC\cong\triangle BAD\) and \(\triangle ABC\cong\triangle CEA\), we have \(\angle ADE=\angle ABC\) and \(\angle AED=\angle ACB\) (corresponding angles of congruent triangles)

By the converse of the corresponding - angles postulate: If two lines are cut by a transversal and the corresponding angles are equal, then the lines are parallel.

Answer:

Since \(\angle DAB+\angle BAC+\angle CAE = 180^{\circ}\), by the definition of a straight - angle (an angle whose measure is \(180^{\circ}\)), points \(D\), \(A\), and \(E\) are collinear.

b.