Question

note : figure not drawn to scale
in the figure, triangle abf is congruent to triangle edg, where b corresponds to d. the measure of angle bcg is 56. what is the measure, in degrees, of angle edg?
a. 28
b. 34
c. 56
d. 62

Explanation:

Step1: Identify right angles

Triangles \(ABF\) and \(EDG\) are right - angled (since \(\angle A\) and \(\angle E\) are right angles, \(90^{\circ}\)). Also, \(\triangle ABF\cong\triangle EDG\), so corresponding angles are equal. \(\angle BCG\) and \(\angle DCF\) are vertical angles, so \(\angle DCF = \angle BCG=56^{\circ}\).

Step2: Use triangle angle sum

In right - triangle \(EDG\), \(\angle E = 90^{\circ}\), and we know that the sum of angles in a triangle is \(180^{\circ}\). Let \(\angle EDG=x\). In \(\triangle EDG\), we also know that \(\angle DCF\) is an exterior angle or we can use the fact that in the right - triangle, \(\angle EDG + \angle EGD=90^{\circ}\), and \(\angle EGD\) is related to the vertical angle. Wait, another approach: Since \(\triangle ABF\cong\triangle EDG\), \(\angle B=\angle D\). In \(\triangle BCG\), \(\angle B + \angle BCG=90^{\circ}\) (because \(\angle A = 90^{\circ}\) and \(AB\parallel EG\) maybe? Wait, more simply, in right - triangle \(ABF\), \(\angle B + \angle BFA = 90^{\circ}\), and in right - triangle \(EDG\), \(\angle D+\angle DGE = 90^{\circ}\). But since \(\triangle ABF\cong\triangle EDG\), \(\angle B=\angle D\) and \(\angle BFA=\angle DGE\). Also, \(\angle BCG\) and \(\angle B\) are related. Wait, \(\angle BCG = 56^{\circ}\), and in right - triangle \(ABG\) (wait, \(\angle A = 90^{\circ}\)), so \(\angle B + \angle BCG=90^{\circ}\)? No, \(\angle BCG\) is an angle in the intersection. Wait, the correct way: Since \(\angle A\) and \(\angle E\) are right angles (\(90^{\circ}\)), and \(\triangle ABF\cong\triangle EDG\), so \(\angle B=\angle D\). In \(\triangle BCG\), \(\angle BCG = 56^{\circ}\), and \(\angle B + \angle BCG=90^{\circ}\) (because \(\angle A = 90^{\circ}\) and \(AB\) is perpendicular to \(AG\), so \(AB\parallel EF\) (since \(\angle A=\angle E = 90^{\circ}\)), so \(\angle B+\angle BCG = 90^{\circ}\)). So \(\angle B=90^{\circ}- 56^{\circ}=34^{\circ}\). Since \(\triangle ABF\cong\triangle EDG\), \(\angle D=\angle B = 34^{\circ}\), so \(\angle EDG = 34^{\circ}\).

Answer:

b. 34