Let $ABC$ be an acute triangle inscribed on the circumference $\Gamma$. Let $D$ and $E$ be points on $\Gamma$ such that $AD$ is perpendicular to $BC$ and $AE$ is diameter. Let $F$ be the intersection between $AE$ and $BC$. Prove that, if $\angle DAC = 2\angle DAB$, then $DE = CF$.
The well-known fact $\angle BAH=\angle CAO$ implies that $\angle DAE=\angle CAE,$ so by Thales and AAS we get $\triangle DAE\cong\triangle CAE$ which implies $DE=CE.$ Then note that $$\angle EFC=\angle BFA=90-\angle DAF=90-\angle CAF=\angle FEC$$so $\triangle EFC$ is isosceles and therefore $CE=CF.$
I did not understand this solution well, it is missing some more information. And who is H?
$H$ is the orthocenter of $ABC$, and $O$ is the circumcenter of $ABC$.