Given: \(\angle A \cong \angle D, A E \cong D E, A C \cong \underline{D B} .\) Prove: \(\Delta A B E \cong \triangle D C E\).

Prove that any point on the perpendicular bisector of a line segment is equidistant from the endpoints of the segment.

\(\underline{A D}\) is a straight line segment. Triangles \(A E B\) and \(D F C\) are drawn in such a way that \(\angle \mathrm{A} \cong \angle \mathrm{D}\) and \(\underline{\mathrm{AE}} \cong \underline{\mathrm{DF}}\). Additionally, it is given that \(\underline{A C} \cong \underline{B D}\). Prove that $\Delta A E B \cong \triangle D F C$.

Given: \(\angle 1 \cong \angle 2 ; \angle 3 \cong \angle 4\). Prove $\triangle \mathrm{ADC} \cong \triangle \mathrm{BEC}$.

Given: $\underline{D C} \cong \underline{A B} \cdot \angle C D A \cong \angle B A D .\( Prove: \)\underline{A C} \cong \underline{B D}$.

When looking at the reflection of an object in the mirror, it seems as if the object is actually behind the mirror. The place where the object seems to be is called the virtual image. In the accompanying figure, light from the object at point \(O\) reflects off the mirror \(\underline{M N}\) at point \(B\) and strikes the eye at point \(D\). Given (by the laws of physics) that the angle of incidence equals the angle of reflection (\angleOBM $\cong \angle \mathrm{DBN}$ ) and that the total distance from the eyes to the virtual image, I, must equal the total distance traveled by the light from the eyes to the object, show that the virtual image is diametrically opposite the object with respect to the mirror.