Typo?...

I think $BH\parallel CD$

fura3334 wrote: I think $BH\parallel CD$ Thanks

Should've been USAMO 5

Add midpoint $M$ of $CD$. $BHMC$ is a parallelogram.

Complex bash with quadrilateral $ABCD$ inscribed in the unit circle, so that $$|a|=|b|=|c|=|d|=1$$Since $\angle BAD = 2\angle ADC$ as directed angles, we have $$\frac{b}d = \left(\frac{a}c\right)^2$$so $$d = \frac{bc^2}{a^2}$$Since $CD = 2BC$, we have $$|c-d| = 2|b-c|$$$$|a^2-bc| = 2|b-c|$$$$(a^2-bc)^2 = 4a^2(b-c)^2$$$$\frac{a^2-bc}{a(b-c)} = \pm 2$$$$\frac{a-p}{b-c} = \pm 2$$where $P$ is the point on the unit circle such that $AP\parallel BC$. Now $\angle ABC$ is obtuse (because $\angle ADC$ is acute and they sum to $180^{\circ}$) so $B, P$ lie on opposite sides of $AC$ and the vectors $\overrightarrow{PA}$, $\overrightarrow{CB}$ are pointing in the same direction. So $$\frac{a-p}{b-c} = 2$$and so $$a^2-bc = 2ab-2ac$$$$b = \frac{a(a+2c)}{2a+c}$$$$d = \frac{c^2(a+2c)}{a(2a+c)}$$Then $$h = \frac{a+d+c-ad\overline{c}}2 = \frac{a^3+a^2c+c^3}{a(2a+c)}$$and so $$b-h = \frac{a^2c-c^3}{a(2a+c)} = \frac{c(a+c)(a-c)}{a(2a+c)}$$Then $$c-d = \frac{2c(a+c)(a-c)}{a(2a+c)}$$so $$\frac{b-h}{c-d} = \frac12$$which is real. $\blacksquare$

Simple: Let $M$ be the intersection of $CH$ and $\omega$ (the circumcircle). Also let $E$ be the intersection of $AD$ and $BC$. Let $\angle EDC = x$. Then \[ \angle DCM = 90-x \]so \[\angle ECM = 180-2x-\angle DCM = 90-x\]so CM is the angle bisector of $\angle CDE$ with $H$ as the midpoint of $ED$ and $CDE$ is iscoceles with $CD=CE$. Thus, if $DC=2a$, $EB=BC=a$. Thus, $H$ is the midpoint of $ED$ and $B$ is the midpoint of $EC$ so $HB||DC$.