Question 1025582
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In isosceles triangle AB$ (with AB = AC), point D lies on AB such that CD = CB. If angle ADC = 115 degrees, what is angle ACD (in degrees)?
 image: http://classroom.artofproblemsolving.com/Classes/IntroGeom/Images/480402815.gif
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1.  angle(CDB) = 180° - angle(ADC) = 180° - 115° = 65° (as a complementary angle).


2.  angle(CBD) = angle(CDB) = 65° as the triangle CBD is isosceles.


3.  angle(ACB) = angle(ABC) = 65° as the triangle ABC is isoceles.


4.  From the triangle ABC, angle(A) = 180° - (65°+65°) = 50°.


5.  From the triangle ADC, angle(ACD) = 180° - (115°+50°) = 15°.


The problem is solved.
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