Question

what is the value of t?

Explanation:

Step1: Use the property of congruent triangles

Since \(CF = CD\) (marked with one tick), \(FE = DE\) (marked with two ticks), and \(CE=CE\) (common side), \(\triangle CFE\cong\triangle CDE\) (by SSS - Side - Side - Side congruence criterion).

Step2: Use the property of congruent angles

Corresponding parts of congruent triangles are equal. So, \(\angle F=\angle D\). Also, in \(\triangle CFE\), the sum of angles in a triangle is \(180^{\circ}\). But we can also use the fact that for quadrilateral - related (since we have two congruent triangles forming a figure where we can use angle sum). However, using the triangle congruence directly: \(\angle F = 106^{\circ}\) (given), and from \(\triangle CFE\cong\triangle CDE\), \(\angle D=t\). Another way: consider the sum of angles in the two - triangle figure. But more simply, since \(\triangle CFE\cong\triangle CDE\), \(\angle D=\angle F-(2\times44^{\circ})\) is wrong. Wait, no, correct approach:
The sum of angles in \(\triangle CFE\) and \(\triangle CDE\): but better, since \(\triangle CFE\cong\triangle CDE\), \(\angle FCE=\angle DCE = 44^{\circ}\), \(\angle F=\angle D\). Using the angle - sum property of a triangle in \(\triangle CFE\): \(\angle FCE + \angle FEC+\angle F=180^{\circ}\), and in \(\triangle CDE\): \(\angle DCE+\angle DEC+\angle D = 180^{\circ}\). But since \(\triangle CFE\cong\triangle CDE\), \(\angle F=\angle D\). Also, we can use the fact that if we consider the two triangles together. Wait, the correct formula is \(t = 180^{\circ}-(106^{\circ}+ 44^{\circ})\) is wrong. Wait, no:
Since \(\triangle CFE\cong\triangle CDE\), \(\angle F=\angle D\). Wait, no, wait the two triangles: \(\triangle CFE\) and \(\triangle CDE\). The sum of angles around point \(C\) for the two triangles: no. Wait, correct:
We know that \(\triangle CFE\cong\triangle CDE\) (SSS). So, \(\angle F=\angle D\). But wait, no, wait the two triangles: \(\angle FCE = 44^{\circ}\), \(\angle F=106^{\circ}\). In \(\triangle CFE\), \(\angle FEC=180^{\circ}-(106^{\circ}+44^{\circ}) = 30^{\circ}\). And since \(\triangle CFE\cong\triangle CDE\), \(\angle DEC=\angle FEC = 30^{\circ}\), \(\angle DCE=\angle FCE = 44^{\circ}\). Then in \(\triangle CDE\), using the angle - sum property of a triangle (\(\angle DCE+\angle DEC+\angle D=180^{\circ}\)), \(t=\angle D=180^{\circ}-(44^{\circ}+ 30^{\circ})=106^{\circ}\).

Answer:

\(106\)