Let H be the orthocenter of the acute triangle ABC. Extend the altitude AD until it intersects the circumcircle of triangle ABC at F. Show that BH=BF

Question

I have been doing the problem for an hour and have gotten nowhere, can anyone either give hints OR help guide me through the proof. Thank you.

Answer 1

Because if $BK$ is an altitude of our triangle we obtain: $$\measuredangle BFA=\measuredangle BCA=90^{\circ}-\measuredangle DAC=\measuredangle AHK=\measuredangle BHF,$$ which gives $\measuredangle BFH=\measuredangle BHF$ and we are done!