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Saudi Arabian Mathematical Competitions, 2012

Geometry Difficulty 5.2 AIME, harder Prove it Saudi Arabia

Let $ABC$ be an arbitrary triangle. A circle passes through $B$ and $C$ and intersects the lines $AB$ and $AC$ in $D$ and $E$ , respectively. The projections of the points $B$ and $E$ on $CD$ are denoted by $B'$ and $E'$ , respectively. The projections of the points $D$ and $C$ on $BE$ are denoted by $D'$ and $C'$ , respectively.

Prove that the points $B'$ , $D'$ , $E'$ and $C'$ lie on the same circle.

Solutions — 2

Solution 1

Let $I$ be the intersection point of the lines $BE$ and $CD$ . The quadrilaterals $BD'B'D$ and $CE'C'E$ are cyclic, so $\overline{BDB'} = \overline{B'D'I}$ and $\overline{CEC'} = \overline{IE'C'}$ . Since $BDEC$ is also cyclic, $\overline{BDB'} = \overline{CEC'}$ . It follows that $\overline{B'D'I} = \overline{IE'C'}$ , so $B'D'E'C'$ is a cyclic quadrilateral.

Solution 2

Using the power of a point theorem, one has:
$IB' \cdot ID = ID' \cdot IB$
$IC' \cdot IE = IE' \cdot IC$
$IE \cdot IB = ID \cdot IC$
From these one easily obtains
$IB' \cdot IE' = ID' \cdot IC'$
which proves that the quadrilateral $B'D'E'C'$ is cyclic, using the reciprocal of the power of a point theorem.

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