Questions: Given: triangle ABC is congruent to triangle EFD, find the measure of angle 1.

Solution Steps

Triangle ABC is isosceles with AB = BC. We are given that $\angle B = 68^\circ$. Since the sum of the angles in a triangle is $180^\circ$, and the base angles of an isosceles triangle are equal, we have: $2\angle C + \angle B = 180^\circ$ $2\angle C + 68^\circ = 180^\circ$ $2\angle C = 180^\circ - 68^\circ$ $2\angle C = 112^\circ$ $\angle C = \frac{112^\circ}{2}$ $\angle C = 56^\circ$

Since $\triangle ABC \cong \triangle EFD$, corresponding angles are congruent. Therefore, $\angle C \cong \angle D$. Thus, $\angle D = 56^\circ$.

$\angle 1$ is an exterior angle of $\triangle DEF$. The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles. Therefore, $\angle 1 = \angle D + \angle F$ Since we don't have the measure of $\angle F$, we use the fact that $\triangle ABC \cong \triangle EFD$. Thus, $\angle A \cong \angle E$ and $\angle B \cong \angle F$. We know that $\angle B = 68^\circ$, so $\angle F = 68^\circ$.
Now, we have: $\angle 1 = \angle D + \angle F$ $\angle 1 = 56^\circ + 68^\circ$ $\angle 1 = 124^\circ$

Final Answer