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Singapore International Mathematical Olympiad Committee National Team Selection Test

Geometry Difficulty 7.3 National Olympiad, round 2 Prove it Singapore

In an isosceles triangle $ABC$ , $AB = AC$ . Let $E$ be a point on $AC$ and extend $AB$ to a point $D$ such that $BD = EC$ . Let $F$ be the intersection of $BC$ and $DE$ and let the circumcircle of $ABC$ intersect the circumcircle of $BDF$ at $G$ . Prove that $DE$ is perpendicular to $FG$ .

Solution

Note that the points $B$ , $D$ , $G$ , $F$ are concyclic and the points $A$ , $B$ , $G$ , $C$ are also concyclic. Also since $\angle EDG = \angle FDG = \angle FBG = \angle CBG = \angle GAC = \angle GAE$ , the points $A$ , $D$ , $G$ , $E$ are concyclic. Since $\angle GEC = \angle ADG = \angle BDG = \angle CFG$ , the points $G$ , $C$ , $E$ , $F$ are concyclic. Since $BD = EC$ , $\angle CFE = \angle BFD$ , the circle passing through the points $B$ , $D$ , $G$ , $F$ and the circle passing through the points $G$ , $C$ , $E$ , $F$ are equivalent. And since the points $A$ , $B$ , $G$ , $C$ are concyclic, $\angle DBG = \angle ECG$ and therefore $DG = GE$ . Finally, since the points $B$ , $D$ , $G$ , $F$ are concyclic, $\angle ABC = \angle ABF = \angle FGD$ . Together with $\angle ECF = \angle ACB = \angle ABC$ , we have $\angle FGD = \angle ECF$ . Therefore $DF = FE$ . Hence $DGF \equiv EGF$ , so $\angle DFG = \angle EFG = 90^\circ$ . Thus $DE \perp FG$ .

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