[Math] Prove vertex-orthocenter distance is twice of side_midpoint-circumcenter distance

Question

$AL$ , $BM$ , $CN$ are altitudes , $TD$ , $RF$ , $ES$ are perpendicular bisectors of sides.

How to prove $AQ = 2PD$ ?

By similar triangles $\triangle AQN \sim \triangle CQL$ and $ \triangle CQL \sim \triangle RPD$, all I got was $$\frac {AQ} {PD} = \frac {AN}{RD}$$

Also by basic proportionality theorem for $\triangle CNB$ with segment $QL$ and similar triangles $ \triangle CQL \sim \triangle RPD$ I get $$ \frac {PR} {PD} = \frac {QN} {BL} $$

I have no idea how to proceed further.

Answer 1

the easy way is to have $BP$ extended to $Y$, now prove $AQ=YC$

$AQ$ // $YC$,if we can prove $AY$// $CQ$, the problem is solved.

check $\angle YAC $and $\angle ACN$

$\angle ACN +\angle CAN=?$ $\angle YAC +\angle CAN=?$

you should get answer now.