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, 2011

Geometry Difficulty 7.5 National Olympiad, round 2 Prove it Baltic Way

Let $ABCD$ be a convex quadrilateral such that $\angle ADB = \angle BDC$ . Suppose that a point $E$ on the side $AD$ satisfies the equality
$AE \cdot ED + BE^2 = CD \cdot AE.$
Show that $\angle EBA = \angle DCB$ .

Solution

Let $F$ be the point symmetric to $E$ with respect to the line $DB$ . Then the equality $\angle ADB = \angle BDC$ shows that $F$ lies on the line $DC$ , on the same side of $D$ as $C$ . Moreover, we have $AE \cdot ED < CD \cdot AE$ , or $FD = ED < CD$ , so in fact $F$ lies on the segment $DC$ .

Note now that triangles $DEB$ and $DFB$ are congruent (symmetric with respect to the line $DB$ ), so $\angle AEB = \angle BFC$ . Also, we have
$BE^2 = CD \cdot AE - AE \cdot ED = AE \cdot (CD - ED) = AE \cdot (CD - FD) = AE \cdot CF.$
Therefore
$\frac{BE}{AE} = \frac{CF}{BE} = \frac{CF}{BF}.$
This shows that the triangles $BEA$ and $CFB$ are similar, which gives $\angle EBA = \angle FCB = \angle DCB$ , as desired.

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