Let
be an acute triangle with
the feet of the altitudes lying on
respectively. One of the intersection points of the line
and the circumcircle is
The lines
and
meet at point
Prove that
Proposed by Christopher Bradley, United Kingdom
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Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$
Proposed by Christopher Bradley, United Kingdom
Školjka