Question #fbf51

1 Answer
Jan 31, 2018

#angleDEC=40^@#

Explanation:

So1 :
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Given that #AC# is the diameter, #=> angleABE=90^@#,
given #angleCBE=50^@, => angleEBA=90-50=40^@#
recall that angles subtended by the same arc at the circumference are equal, as #angleECA and angleEBA# are subtended by the same arc #AE, => angleECA=angleEBA=40^@#,
given #AC# // #ED#,
#=> angleDEC=angleECA=40^@#

Sol 2 :
enter image source here
As #AC# is the diameter, #=> angleABC=90^@#,
given #angleCBE=50^@, => angleABE=90+50=140^@#,
opposites angles in a cyclic quadrilateral add up to #180^@#,
as #ABEC# is a cyclic quadrilateral, #=> angleABE+ACE=180^@#,
#=> angleACE=180-angleABE=180-140=40^@#
given #ED# // #AC, => angleDEC=angleACE=40^@#