Step 1: Under stand the question and observe the given figure

Step 2: Recall properties of isosceles triangle

              NOTE: Isosceles triangle has two equal sides and two equal angles.

Step 3: Find all the unknown interior angles of the isosceles triangles ABC

            Skill 1: Know about  the interior angle of isosceles triangles.

                        NOTE: Sum of the interior angles of the isosceles triangle =

                                    2 * (equal angle) + angle = 180 \degree

                                       we have two equal angle.

            Skill 2: Set up an equation equating 180 \degree

                        EXAMPLE: \triangle ABC is an isosceles triangle and  

                                        therefore  \angle CAB = \angle ABC

                                [math] \angle CAB + \angle ABC + \angle ACB = 180 \degree [/math]

            Skill 3: Substitute the known angle and simplify for the unknown angles

                    EXAMPLE:  [math] 2 (\angle CAB or \angle ABC) = 180 \degree - 66 \degree[/math]

                                   [math] 2 (\angle CAB or \angle ABC) = 113 \degree [/math]

                              \angle CAB = \angle ABC = [math] \frac{113}{2} = 57 \degree[/math]

Step 4: Follow the same process to find the unknown angles of right isosceles triangles(BCD)

            NOTE: It is a right  isosceles triangles(BCD) \angle BCD = 90

          EXAMPLE: \angle CBD = \angle CDB = \frac{180 - 90}{2}

                                [math] \angle CBD = \angle CDB = 45 \degree [/math]

Step 5: Find the all angles of the right \triangle BDE

              1) \triangle BDE    is a right angle so \angle BED = 90

              2) Find the \angle DBE =  ?

             NOTE: \angle ABC, \angle CBD and \angle DBE make a straight line

                       Hence   \angle ABC  + \angle CBD +   \angle DBE  = 180

                           Gives   \angle DBE   = 180 \degree  - ( \angle ABC  + \angle CBD )

                                           \angle DBE = 180 \degree  - 102 \degree  = 78 \degree

                3)  Find  the \angle BDE   

                     NOTE: Sum of the angles in triangle = 180

                                So   \angle BED +   \angle DBE + \angle BDE = 180

                                       \angle BDE   = 180 \degree  - (90 \degree  + 78 \degree)

                                         \angle BDE = 12 \degree