A RIGHT TRIANGLE and ... one more // Un TRIUNGHI DREPTUNGHIC și ... încă unul // Правоаголен триаголник и ... уште еден

 It's a fairly basic geometry problem (proposed for sixth grade). 

    "E :  6209*.  In triangle  $ABC$  with  $\sphericalangle A=90^{\circ}$ the bisector of angle  $C$ intersects the perpendicular bisector of side  $AB$  in  $D$. Find  $\sphericalangle  CBD$.

 Ion MACREA, professor Brașov"

                    Figure of the problem (CiP) : 

               REMARK CiP  The order of important lines in a triangle is known to be: altitude -  - - bisectors - median - perpendicular bisectors. Since CF is the median and CA the altitude, line CD is between them, so point D is on the extension of segment EF, beyond F and not elsewhere.

ANSWER CiP

$\sphericalangle CDB=90^{\circ}$

                    Solution CiP 

               We rely on the following characterization of right triangles :

                           In a right triangle, the median to the hypotenuse

                          is equal to one half of the hypotenuse.

              Conversely, if in a triangle the median drawn from a vertex

                      has length equal to one half of the side to which it is drawn,

                       then the angle at that vertex is a right angle.

A complete figure is below. Point  $F$  is the midpoint of the leg  $[AB],\;\;DE\perp AB$.

(Basically,  $EF$  is the perpendicular bisector of side  $[AB]$.) 

          $FE,\;AC \perp AB \Rightarrow FE \parallel AC$  and then, 

once

 E is the midpoint of the segment BC                         (1)

 and the second time we have the equalities of angles

$\sphericalangle EDC=\sphericalangle ECD \;\;\;\overset{CD\; is\; the}{\underset{bisector\; of \;\sphericalangle ACB}{=}}\;\;\;\sphericalangle ACD\tag{2}$

             $(2)\Rightarrow\;\;CDE$  is isosceles triangle, with base  $ CD$  so

$DE=EC\underset{(1)}{=}EB=\frac{BC}{2}$

but this says that, in triangle  $BCD$  

    the median  $DE$  drawn from the vertex  $D$  has length equal to one half of the side  $BC$  to which it is drawn,

so  the angle  $\sphericalangle BDC$  at the vertex  $D$  is a right angle.

$\blacksquare$