Altitudes

[Fatin Farhan]

Let $$ABC$$ be an acute angle triangle with $$D,E,F$$ the feet of altitude line of $$BC,CA,AB$$ respectively. One of the intersection points of the EF and the circumcircle is P. BP and DF meet at point Q. Show that, $$AP=AQ$$

[asif e elahi]

$\angle APQ=\angle APB=\angle ACB=\angle BFD=\angle BFQ=180^{\circ}-\angle AFQ$.So $APQF$ cyclic.
Again,$\angle AEP=\angle CEF=180^{\circ}-CBF=180^{\circ}-ABC=APC$.So $\bigtriangleup AEP\sim \bigtriangleup APC$.
$\frac{AP}{AE}=\frac{AC}{AP}$
or $AP^{2}=AE.AC=AF.AB$
or $\frac{AP}{AF}=\frac{AB}{AP}$
So $\bigtriangleup AFP\sim \bigtriangleup ABP$
$\angle ABP=\angle APF$
or $\angle ABQ=\angle AQF$
So $\bigtriangleup AQF\sim \bigtriangleup ABQ$
$\frac{AQ}{AF}=\frac{AB}{AQ}$
So $AQ^{2}=AF.AB=AE.AC=AP^{2}$

[Fatin Farhan]

My solution was a bit small
$$\angle APQ=\angle C=\angle BFD=180^{\circ}-\angle AFQ$$.So $$APQF$$ cyclic.
$$\angle AQP=\angle AFP= \angle C$$
$$\angle APQ= \angle AQP$$ $$=>AQ=AP$$