Given a triangle ABC, let $B'$ and $C'$ be points on the sides AB and AC such that $BB' = CC'$. Let $O$ and $O'$ be the circumcentres (i.e. the centre of the circumscribed circle) of triangles ABC and $AB'C'$, respectively.Suppose $OO'$ intersect lines $AB'$ and $AC'$ at $B''$ and $C''$, respectively. If $AB = 1/2 AC$, then
A) $AB'' < 1/2 AC''$ B) $AB'' = 1/2 AC''$ C) $1/2 AC'' < AB'' < AC''$ D) $AB'' = AC''$ E) $AB'' > AC''$
Since we have circumscribed circles, I dropped down perpendiculars from $O'$ and $O$ to sides $AB$ (and $AB'$) and $AC$ (and $AC'$) so that I have right triangles. Also, I know that perpendiculars should divide the side of the triangle in half (so a perpendicular from $O'$ to side $AB'$ would bisect $AB'$, for example. But I'm not sure what to do after that.
Let \begin{align} |BB'|&=|CC'|=u ,\\ \angle BAC&=\alpha ,\quad \angle CBA=\beta ,\quad \angle ACB=\gamma ,\\ |AB|&=c,\quad |AC|=b=2c ,\quad |BC|=a \\ |AB'|&=c-u,\quad |AC'|=b-u=2c-u ,\quad |B'C'|=a' ,\\ \sin\beta&=2\sin\gamma ,\\ |OA|=|OB|=|OC|&=R =\frac{c}{2\sin\gamma} ,\\ M_b&=\tfrac12(A+C) ,\\ M_b'&=\tfrac12(A+C') ,\\ |AO'|=|CO'|&=R' . \end{align}
\begin{align} \angle COA&=2\beta ,\\ \angle COM_b&=\beta ,\\ \angle ACO =\angle OAC &=\tfrac\pi2-\beta \end{align}
\begin{align} a&= c\sqrt{5-4\cos\alpha} ,\\ R&=\frac{c\sqrt{5-4\cos\alpha}}{2\sin\alpha} ,\\ |OM_b|&=R\sin\angle ACO =R\cos\beta \\ &= \frac{c\sqrt{5-4\cos\alpha}}{2\sin\alpha} \cdot \frac{a^2+c^2-b^2}{2ac} \\ &= \frac{c\sqrt{5-4\cos\alpha}}{2\sin\alpha} \cdot \frac{c^2(5-4\cos\alpha)+c^2-4c^2}{2c^2\sqrt{5-4\cos\alpha}} \\ |OM_b|&=\frac{c(1-2\cos\alpha)}{2\sin\alpha} . \end{align}
\begin{align} a'&=\sqrt{b'^2+c'^2-2b'c'\cos\alpha} \\ &= \sqrt{5c^2-6cu+2u^2-2(2c-u)(c-u)\cos\alpha} ,\\ R'&=\frac{a'}{2\sin\alpha} ,\\ |O'M_b'|^2&=R'^2-|M_b'C'|^2 =\frac{a'^2}{4\sin^2\alpha} - (c-\tfrac12u)^2 \\ &= \frac{((2c-u)\cos\alpha+u-c)^2}{4\sin^2\alpha} ,\\ |O'M_b'|&= \frac{c-u-(2c-u)\cos\alpha}{2\sin\alpha} . \end{align}
\begin{align} \tan\angle OC''C &= \frac{|OM_b|-|O'M_b'|}{|AM_b|-|AM_b'|} = \frac{ \frac{\tfrac12(1-\cos\alpha)\,u}{\sin\alpha} }{\tfrac12u} \\ &= \frac{1-\cos\alpha}{\sin\alpha} =\tan\tfrac\alpha2 . \end{align}
Hence, in $\triangle AB''C''$ we have \begin{align} \angle C''AB''&=\pi-\alpha ,\\ \angle B''C''A&=\tfrac\alpha2 ,\\ \angle AB''C''&=\pi -(\pi-\alpha+\tfrac\alpha2) =\tfrac\alpha2 , \end{align}
hence, $|AB''|=|AC''|$ and the answer is "D".