BdMO Online Forum

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 United Kingdom

$\angle APQ=\angle BFD=\angle C$.
so quadrilateral $AFPQ$ is cyclic.\
so $\angle AFP= \angle AQP=\angle C$
hence $\angle AQP=\angle APQ $
so $AP=AQ$

wouldn't there be another case.................??? @Tahmid

fermat's rebirth wrote:wouldn't there be another case.................??? @Tahmid

yes,i'm working on that.

Complete Solution:

$\angle AGB = 180 - \angle C$ So $\angle AGI =\angle C$
And $\angle AFI = \angle C $ so $AFGI$ cyclic.
So $\angle AIG =\angle C =\angle AGI$ and $AG=AI$
Again,
$\angle AHB =\angle C$ and $\angle AFD =180-\angle C$ so $AFJH$ cyclic.
Now, as $\angle AJH = \angle AFE = \angle C$, $AJ=AH$.