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Because I couldn't get any help on the last problem I got a whole bunch more. Could someone please help with this?

Let AB be a diameter of a circle centered at O. Let E be a point on the circle, and let the tangent at B intersect the tangent at E and AE at C and D, respectively. If angle BAE = 43 degrees, find angle CED, in degrees.

AnonymousConfusedGuy Apr 27, 2018

#1**+1 **

Angle BAE = 43° ...so minor arc BE will = 86°

And angle CBE = 1/2 of this arc = 43°

But.......since CB and CE are tangents to the circle from a common point C....they are equal

Thus...in triangle BCE....BC = EC...and the angles opposite these sides are also equal...

So angle CBE = angle BEC

And note that in triangle AEB.....angle AEB intercepts a diameter....so its measure = 90°

But angle AEB + angle BEC + angle CED = 180°.....so....

90 + 43 + angle CED = 180

133 + angle CED = 180 subtract 133 from each side

angle CED = 47°

CPhill Apr 27, 2018