How to prove this circumcircle question If H is the orthocentre of triangle ABC and AH meets BC at D and the circumcircle at E,then prove that HD=DE

Take a look at triangles $\triangle CDH$ and $\triangle CDE$. 
$\angle CDH=\angle CDE=90^\circ\tag{1}$
$\angle HCD=\angle BAE$ because these two angles have perpendicular legs.
On the other side $\angle BAE=\angle BCE$ because these are inscribed angles over the same arc. Consequentially: 
$\angle HCD=\angle BCE=\angle DCE\tag{2}$ 
Taking (1) and (2) into account, triangles $\triangle CDH$ and $\triangle CDE$ are congruent by SAS. 
It follows immediately that $DH=DE$.
