**Solution.** Denote by *B*' and *D*' the perpendicular foots of points *B* and *D* on plane *ACE*, respectively. Since the angles between plane *ACE* and planes *ABC*, *ABE* and *BCE* are all 45^{o}, the distances between *B*' and lines *AC*, *AE* and *CE* are all equal to *BB*', independently of the position of *B*'. Similarly, the distances between *D*' and lines *AC*, *AE* és *CE* are all equal to *DD*' (Figure 1).

Figure 1.

There exists four such points in the plane of triangle *ACE* which have equal distances from the three lines *AC*, *AE*, *CE*: the incenter and the three excenters. Denote them by *I* and, using the usual indexing, by *I*_{A}, *I*_{C} and *I*_{E}, respectively. Points *B*' and *D*' are two of these four.

Planes *ABE* and *ADE* are symmetric to plane *ACE* and also planes *BCE* and *CDE* are symmetric; this implies that lines *BE* and *DE* are symmetric to plane *ACE*. The perpendicular projections of these two lines coincide. Therefore, points *B*' and *D*' lie on the same (interior or exterior) bisector of angle *AEC*.

Figure 2.

If *B*' and *D*' lie on the interior bisector, then they are *I* and *I*_{E}. Due to the symmetry, it can be assumed that *B*'=*I* and *D*'=*I*_{E} (Figure 2.). From the right triangles *ABB*', *ADD*' and *AB*'*D*',

*AB*^{2}+*AD*^{2}=(*AB*'^{2}+*BB*'^{2})+(*AD*'^{2}+*DD*'^{2})=

=(*AB*'^{2}+*AD*'^{2})+*BB*'^{2}+*DD*'^{2}=*B*'*D*'^{2}+*BB*'^{2}+*DD*'^{2}.

Replacing *A* by *C*,

*BC*^{2}+*CD*^{2}=(*B*'*C*^{2}+*BB*'^{2})+(*CD*'^{2}+*DD*'^{2})=

=(*B*'*C*^{2}+*CD*'^{2})+*BB*'^{2}+*DD*'^{2}=*B*'*D*'^{2}+*BB*'^{2}+*DD*'^{2}.

Therefore *AB*^{2}+*AD*^{2}=*BC*^{2}+*CD*^{2}, the statement holds.

3. ábra

If *B*' and *D*' lie on the exterior bisector of angle *AEC* then they are *I*_{A} and *I*_{C}. The quadrilateral *AB*'*CD*', which is the projection of face *ABCD* is sel-intersecting (Figure 3). This case is not possible.