Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National Olympiad Prove it Italy

Problem:

Let $ABC$ be an isosceles triangle on base $BC$ and let $D, E$ be points on the sides $AB, BC$ respectively, such that the lines $DE$ and $AC$ are parallel. Consider also the point $F$ on the line $DE$ which lies on the opposite side of $D$ with respect to $E$ and is such that $FE$ is congruent to $AD$ . Letting $O$ be the circumcenter of triangle $BDE$ , prove that the points $O, F, A, D$ lie on a circle.

Solution

Solution:

We note that triangle $BDE$ is isosceles on base $BE$ . Indeed, considering the parallel segments $ED, CA$ and the segment $CB$ , we have $\widehat{ACB} = \widehat{DEB}$ . Then, since $\widehat{ACB} = \widehat{CBA}$ , we obtain $\widehat{DEB} = \widehat{CBA}$ .

Let us now consider triangles $OED$ and $ODB$ : they are congruent because $ED = DB$ , $OE = OD$ , $OD = OB$ . In particular, $\widehat{DEO} = \widehat{BDO}$ .

Finally, let us look at triangles $FEO$ and $ADO$ : they have $\widehat{OEF} = \widehat{ODA}$ (because they are supplementary to congruent angles), $OE = OD$ and $EF = AD$ . Hence the two triangles are congruent, and in particular $\widehat{EFO} = \widehat{DAO}$ , from which $\widehat{DFO} = \widehat{DAO}$ . Since $F$ and $A$ lie on the same side of the line $DO$ , this shows that the quadrilateral $O F A D$ is inscribable in a circle.

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