Question

Q. Let ABC be a triangle and D be the mid-point of side BC. Suppose ∠DAB = ∠BCA
and ∠\DAC = 15◦. Show that ∠ADC is obtuse. Further, if O is the circumcentre of
ADC, prove that triangle AOD is equilateral. ........ Take your time |||| 35 pts for 1st come 1st served ANSWERS....... 

Answer 1

consider the attached figure,
from ΔABD, AB/BD=sin(15+x)/sinx    (sine rule)
from ΔABC, AB/BC=sin(x)/sin(15+x)    (sine rule)

hence, 2sin²x=sin²(15+x)
(+/-)√2sinx= sin(15+x)=sin15 cosx+cos15 sinx

cotx=(((+/-)√2)-cos15)/sin15

since cos 15° = (√3+1)/2√2
sin 15° = (√3−1)/2√2

so cotx=√3 =>x=30°
or cotx=-(√3+5)/(√3-1)=-(4+3√3)
since x is not negative x≈180-6.2°=173.8°
since (15+x) is an angle of ΔABD
cotx$\neq$-(4+3√3)

 hence ∠ADC= 135°
 hence obtuse

 let ∠ADO=∅ 
AD/sin30°=DC/sin15°
AD/2=ODcos∅
DC/2=ODcos(135-∅)

solving all 3....
tan∅=√3
∅=60

∠AOD=2*(180-90-60)=60

since all ∠s of the ΔAOD is 60°

ΔAOD is equilateral