We have $\angle BAD=30^\circ$. Consider the (unique) circle $\omega$ through $A$ and $D$ tangent to $BA$, let it cut line $BD$ at $D$ and $X$. Then by inscribed angles, $\angle BXA=\angle DAB=30^\circ$. Conversely, if $\angle BXA=30^\circ$, then $BD$ is exactly the line tangent to circle $ADC$. Hence $\angle BCA=30^\circ$ if and only if circle $ADC$ is tangent to $BA$. By Power of a Point, this is equivalent to $BA^2=BD\cdot BC$. By the Law of Sines, $\frac{BA}{BD}=\frac{\sin 45^\circ}{\sin 30^\circ}=\frac{\sqrt 2/2}{1/2}=\sqrt 2$, so this is equivalent to $2\cdot BD^2=BD\cdot BC$, i.e. $BC=2\cdot BD$. Done.

Rather disappointing geometry problem for #3 in a national olympiad

john111111 wrote: Rather disappointing geometry problem for #3 in a national olympiad yea, very trivial question , they shouldnt have asked this question in national olympiad also is this senior level??

Yes,this was a senior's problem.This year's problems,in contrary to the 2 previous years ones,were of a very low quality,unfortunately.

We can also use cot rule using AD as a cevian. By Cot rule (m+n)cot ADB = m cot C - n cot B ( where m:n is the ratio BD:CD ) 2 cot 45 = cot C - cot 105 ( Since m=n=1 ) Solving further , we get C = 30 For the converse, we can take C = 30 and find m and n.

Let angle ACD = x angle ADB= 45° angle DAC+ angle ACD= 45° angle DAC= 45°-x In ∆ ADC, by sine rule sin(45°-x)/CD= sin x/AD sin x/sin(45°-x)= AD/BD[ Since , BD= CD ] In ∆ ABD, angle BAD= 30° So, by sine rule sin 30°/BD= sin 105°/AD AD/BD= sin 105°/sin 30° AD/BD=√3+1/√2 sin x/sin(45°-x)=√3+1/√2 So, if we solve then, cot x=3+√3/1+√3 = √3 So, x= 30° Conversely, D is midpoint of BC