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2015-09-15T13:35:50+05:30
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

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2015-09-15T17:22:24+05:30

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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°

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