Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National Olympiad Prove it Serbia

На страницама $BC$ и $AC$ троугла $ABC$ дате су тачке $D$ и $E$ , редом. Нека је $F$ ( $F \neq C$ ) тачка пресека кружнице описане око троугла $CED$ и праве која садржи тачку $C$ и паралелна је са правом $AB$ . Нека је $G$ тачка пресека праве $FD$ и странице $AB$ , а $H$ тачка на правој $AB$ таква да је $\text{HDA = GEB}$ и $H-A-B$ . Ако је $DG = EH$ , доказати да тачка пресека дужи $AD$ и $BE$ припада симетрали угла $ACB$ .
(Милош Милосављевић)

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Problem:

On the sides $BC$ and $AC$ of triangle $ABC$ points $D$ and $E$ are given, respectively. Let $F$ ( $F \neq C$ ) be the point of intersection of the circle circumscribed about triangle $CED$ and the line containing point $C$ that is parallel to line $AB$ . Let $G$ be the point of intersection of line $FD$ and side $AB$ , and let $H$ be a point on line $AB$ such that $\text{HDA = GEB}$ and $H-A-B$ . If $DG = EH$ , prove that the point of intersection of segments $AD$ and $BE$ belongs to the bisector of angle $ACB$ .
(Miloš Milosavljević)

Solution

Solution:

Since $\text{AGD = 180 - CFD = CED = 180 - AED}$ , the points $A$ , $E$ , $D$ , $G$ are concyclic. From this it follows that $\text{DAG = DEG}$ .

However, then $\text{DHB = DAG - HDA = DEG - BEG = DEB}$ , so the points $H$ , $E$ , $D$ , $B$ are also concyclic. It follows that $\text{BHE = 180 - BDE = CDE = CFE}$ , which

means that the points $F$ , $E$ and $H$ are collinear.

Now $EH = EF \cdot \frac{AE}{EC}$ and $DG = DF \cdot \frac{DB}{CD}$ , so the condition $EH = DG$ becomes $\frac{AE}{EC} \cdot \frac{CD}{DB} = \frac{DF}{EF}$ . On the other hand, from $\text{DEF = DCF = ABC}$ and $\text{DFE = ACB}$ it follows that $\triangle DEF \sim \triangle ABC$ , so $\frac{DF}{EF} = \frac{AC}{CB} = \frac{AM}{MB}$ , where $M$ is the intersection point of the bisector of angle $ACB$ and side $AB$ .

Therefore, $\frac{AE}{EC} \cdot \frac{CD}{DB} \cdot \frac{BM}{MA} = 1$ , so by Ceva's theorem the lines $AD$ and $BE$ intersect on $CM$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from sr; metadata (topic, difficulty) added by this project.