ABC and BCD are isosceles triangles. Find the size of angle BDE.

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