In triangle ABC, angle ACB is 50 degrees, and angle CBA is 70 degrees. Let D be the foot of the perpendicular from A to BC, O the center of the circle circumscribed about triangle ABC, and E the other end of the diameter which goes through A. Find the angle DAE, in degrees.
Extend AD through to intersect the circle at F
Now.....since angle ACB = 50°...then minor arc AB has twice its measure =100°
And since angle CBA =70°....then angle BAD = BDA - CBA = 90 - 70 = 20° = angle BAF
And angle BAF = intercepts minor arc BF...so the measure of minor arc BF is twice this = 40°
But arc ABE =180° = minor arc AB + minor arc BF + minor arc FE
180 = 100 + 40 + minor arc FE
180 =140 + minor arc FE
40 = minor arc FE
But angle FAE intercepts minor arc FE, so it has 1/2 of its measure = 20°
But angle FAE = angle DAE...so DAE = 20°
There are 180° in triangle ACD, so..
m∠DAC = 180° - 90° - 50°
m∠DAC = 40°
An inscribed angle is half the measure of its intercepted arc, so....
(1/2)m∠AOC = m∠ABC
(1/2)m∠AOC = 70°
m∠AOC = 140°
OA and OC are radii of circle O , so they are the same length, and triangle AOC is isosceles. So...
m∠OAC = m∠OCA
m∠OAC + m∠OCA + 140° = 180°
m∠OAC + m∠OAC + 140° = 180°
2m∠OAC + 140° = 180°
2m∠OAC = 40°
m∠OAC = 20°
m∠DAE = m∠DAC - m∠OAC
m∠DAE = 40° - 20°
m∠DAE = 20°