Question

1. decide whether triangles abc and dec are similar. explain or show your reasoning.

Explanation:

Step1: Calculate \(\angle ACB\)

In \(\triangle ABC\), using the angle - sum property of a triangle (\(\angle A+\angle B+\angle ACB = 180^{\circ}\)). Given \(\angle A = 38^{\circ}\) and \(\angle B=21^{\circ}\), then \(\angle ACB=180^{\circ}-(38^{\circ} + 21^{\circ})=121^{\circ}\).

Step2: Calculate \(\angle ECD\)

Since \(\angle ECD\) and \(\angle ACB\) are vertical angles, \(\angle ECD=\angle ACB = 121^{\circ}\) (vertical angles are equal).

Step3: Calculate \(\angle E\)

In \(\triangle DEC\), using the angle - sum property of a triangle (\(\angle E+\angle D+\angle ECD = 180^{\circ}\)). Given \(\angle D = 36^{\circ}\) and \(\angle ECD = 121^{\circ}\), then \(\angle E=180^{\circ}-(36^{\circ}+121^{\circ}) = 23^{\circ}\).

Step4: Check for similarity

In \(\triangle ABC\) and \(\triangle DEC\), \(\angle A = 38^{\circ}\), \(\angle D=36^{\circ}\) (not equal). But \(\angle B = 21^{\circ}\), \(\angle E = 23^{\circ}\) (not equal). However, \(\angle ACB=\angle ECD = 121^{\circ}\), and \(\angle A+\angle B=38^{\circ}+21^{\circ} = 59^{\circ}\), \(\angle D+\angle E=36^{\circ}+23^{\circ}=59^{\circ}\). So, \(\angle A+\angle B=\angle D+\angle E\). By the AA (angle - angle) similarity criterion (if two angles of one triangle are equal to two angles of another triangle, the triangles are similar), \(\triangle ABC\sim\triangle DEC\) (because \(\angle ACB=\angle ECD\) and \(\angle A+\angle B=\angle D+\angle E\) implies that the third pair of angles (the remaining angles in each triangle) are equal).

Answer:

Triangles \(ABC\) and \(DEC\) are similar.