Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 5.0 AIME Prove it Czech Republic

Let $ABC$ be an acute triangle with altitude $AD$ . The bisectors of angles $BAD$ , $CAD$ , intersect side $BC$ at $E$ , $F$ , respectively. The circumcircle of triangle $AEF$ intersects sides $AB$ , $AC$ at $G$ , $H$ , respectively. Prove that lines $EH$ , $FG$ , and $AD$ pass through a common point. (Patrik Bak)

Solutions — 2

Solution 1

Let $K$ be the intersection of segments $FG$ and $AE$ and $L$ the intersection of $EH$ and $AF$ (Fig. 2). Inscribed angles give

$\angle AGF = \angle AEF = 90^\circ - \angle DAE = 90^\circ - \angle GAE,$

Fig. 2

that is $\angle AGF + \angle GAE = 90^\circ$ , hence
$\angle AKG = 180^\circ - (\angle AGF + \angle GAE) = 90^\circ.$
Line *FK* is therefore an altitude of the triangle *AEF*. Similarly we prove that *EL* is its altitude too, thus the intersection of *FK* and *EL* is the orthocenter of triangle *AEF* and it lies on its third altitude *AD* too.

Solution 2

Let *M* be the second intersection of ray $AD$ and the circumcircle $k$ of triangle $AEF$ (Fig. 2). Since $AE$ is the bisector of angle $GAM$ , the angles $GAE$ and $EAM$ inscribed in $k$ are equal and therefore the chords $GE$ and $EM$ are also equal. Furthermore, the corresponding inscribed angles $GFE$ and $EFM$ are equal too. The intersection of $FG$ and $AM$ is thus the reflection of $M$ about $EF$ . The same argument applies to the intersection of $EH$ and $AM$ , therefore the original intersections are identical.

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