In the adjoining figure, CD is the diameter of a semicircle with center O. Point A lies on the extension of DC past C; point E lies on the semicircle, and B is the point of intersection (distinct from E) of line segment AE with the semicircle. If length AB equals length OD, and the measure of angle EOD is 45 degrees, then find the measure of angle BAO, in degrees.
∠CED = 90º ∠ADE = 67.5º ∠ACE = 90 + 67.5 = 157.5º
∠BAO + ∠AEC = 180 - 157.5 = 22.5º
∠BAO : ∠AEC = 2 : 1
Angle BAO = 15 degrees
Hey, jugoslav.....could you show how you solved this ???
I couldn't figure out any way to solve it
First, I calculated the angle ADE (see my diagram), and then I subtracted that angle plus 90º angle (∠CDE) from 180, and I got 22.5
Value of 22.5 is the sum of angles BAO and AEC.
The ratio of these 2 angles is ∠BAO : ∠AEC = 2 : 1
And the rest is easy