Mathematical Olympiad questions: Plane Geometry-3

Known: Isosceles trapezoid $ABCD $, $AD \parallel bc$, in $AB $ out square $ABPQ $, in $CD $ out square $CDRS $, $QR $ $AD $ in $E $, $PS $ for $BC $ in $F $.

Verification: $EF \perp bc$.

Analysis:

by isosceles trapezoid and square nature, consider constructing congruent triangles.

Prove:

Over $Q, S $ is $AH $ vertical cross $AH $ (or its extension) in $G, h$.

by $\angle{qga} = \ANGLE{RHD} = 90^{\circ},\ \angle{aqg} = \angle{gab} = \angle{abc} = \ANGLE{DCB} = \angle{drh},\ AQ = AB = CD = dr$,

Can get $\TRIANGLE{AQG} \cong \triangle{drh}$, and then easy to license $\triangle{eqg}\cong \triangle{erh} \rightarrow E$ is $GH $ midpoint.

Similarly, over $P, S $ is made $BC $ perpendicular to the $BC $ (or its extension line) in $I, j$, easy license $F $ is $IJ $ among the points.

$\because \triangle{aqg}\cong\triangle{bpi}\cong\triangle{drh}\cong\triangle{csj}$,

$\therefore$ Quadrilateral $GIJH $ is isosceles trapezoid.

and $E, f$ respectively is the midpoint of the bottom, so $EF \perp bc$.

Q$\cdot$e$\cdot$d

Mathematical Olympiad questions: Plane Geometry-3