sadikabid27 wrote:

No

Bunuel I could not relate the picture with your solution given earlier. Could you please explain it with picture how \(\angle AEB = 180 - 2*\angle EDC = 150^o\)?

and \(\angle AEB = 180 - 2*\angle BAC = 180 - 2*(180 - 90 - \angle ACB) = 150^o\)?

For (1). First of all, \(\angle AEB = \angle DEC\). Next, DEC is isosceles, thus \(\angle EDC = \angle ECD=15^o\). The sum of the angles in a triangle is 180 degrees:

\(\angle EDC + \angle ECD + \angle DEC = 180^o\);

\(\angle EDC + \angle EDC + \angle AEB = 180^o\);

\(\angle AEB = 180 - 2*\angle EDC = 150^o\).

For (2). First of all, \(\angle BAC + \angle ACB= 90^o\) --> \(\angle BAC =90^o-\angle ACB\). Next, AEB is isosceles, thus \(\angle BAC = \angle ABE\). The sum of the angles in a triangle is 180 degrees. \(\angle BAC + \angle ABE + \angle AEB = 180^o\);

\(\angle BAC + \angle BAC + \angle AEB = 180^o\);

\(\angle AEB =180^o-2*\angle BAC = 180^o-2*(90^o-\angle ACB)\)

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