Question

In figure, Angle ACB is right angle & AC=CD & CDEF is Parallelogram .If Angle FEC=10,then Calculate angle BDE...

Answer 1

now triangle ACB is a right triangle.
so,∠acb =90°
as AC II DE, ∠BDE=90°.
As CDEF is a parallelogram,
∠f=∠d.
as AC=CD,∠CAD=∠CDA
THUS,∠ACD=∠BDE

Answer 2

angle ECD = angle FEC = 10°  as CD and EF are parallel.
angle DCA = angle BCA - angle ECD = 90 -10° = 80°

in triangle ACD, AC = CE  . so it is an isosceles triangle.
   so angle CAD = angle ADC  = (180 - angle DCA)/2 = 50°

in triangle DCE, angle CDE = 180 - angle DCE - angle DEC =
         = 180 - 10 - 90 = 80°

Hence  at point D on AB,
  angle BDE = 180 - angle ADC - angle CDE = 180 - 50 - 80 = 50°