#4 Quadrilaterals

Draw diagonal \(AC\). Since \(AB\parallel CD\), we have \(\angle BAC=\angle ACD\). Since \(AD\parallel BC\), we have \(\angle DAC=\angle ACB\). Adding the equal parts gives \(\angle A=\angle C\). Similarly, drawing diagonal \(BD\) gives \(\angle B=\angle D\).

Draw \(AC\). From \(AB\parallel CD\), \(\angle BAC=\angle ACD\); from \(AD\parallel BC\), \(\angle BCA=\angle CAD\). Side \(AC\) is common. Thus \(\triangle ABC\cong\triangle CDA\) by a side and two adjacent angles. Therefore \(AB=CD\) and \(BC=AD\).

Let \(ABCD\) be a rectangle. Consider triangles \(ABC\) and \(DCB\). We have \(AB=CD\) because a rectangle is a parallelogram, side \(BC\) is common, and angles \(ABC\) and \(DCB\) are right. Thus the triangles are congruent, so \(AC=BD\).

In a rhombus all sides are equal, so \(AB=AD\) and \(BC=CD\). Side \(AC\) is common. Hence \(\triangle ABC\cong\triangle ADC\) by SSS. Therefore \(\angle BAC=\angle CAD\) and \(\angle BCA=\angle ACD\). Diagonal \(AC\) bisects angles \(A\) and \(C\).

By the trapezoid midline theorem, \(MN=\frac{AD+BC}{2}=\frac{17+9}{2}=13\).

Draw \(AC\). Since \(AB\parallel CD\), \(\angle BAC=\angle ACD\). Also \(AB=CD\), and \(AC\) is common. Thus \(\triangle BAC\cong\triangle DCA\) by SAS. Therefore \(\angle BCA=\angle CAD\), so \(BC\parallel AD\). Together with \(AB\parallel CD\), this gives a parallelogram.

Triangles \(AOB\) and \(COD\) are congruent by SAS: \(AO=OC\), \(BO=OD\), and angles \(AOB\) and \(COD\) are vertical. Hence \(\angle ABO=\angle CDO\), so \(AB\parallel CD\). Similarly, triangles \(AOD\) and \(COB\) are congruent, so \(AD\parallel BC\). Both pairs of opposite sides are parallel, hence \(ABCD\) is a parallelogram.

Let \(\angle A=90^\circ\) in parallelogram \(ABCD\). Since \(AD\parallel BC\), angles \(A\) and \(B\) are same-side interior angles, so \(\angle A+\angle B=180^\circ\). Hence \(\angle B=90^\circ\). Opposite angles of a parallelogram are equal, so \(\angle C=90^\circ\) and \(\angle D=90^\circ\). All angles are right, so it is a rectangle.

In a parallelogram, \(AB=CD\) and \(BC=AD\). By condition \(AB=BC\). Hence \(AB=BC=CD=AD\). All sides are equal, so \(ABCD\) is a rhombus.

Let \(MN\) be the midline. Then \(14=\frac{AD+8}{2}\). Hence \(AD+8=28\), so \(AD=20\).

Since \(AB=CD\), the trapezoid is isosceles, so \(\angle DAB=\angle CDA\). From \(AD\parallel BC\), we get \(\angle DAB+\angle ABC=180^\circ\). Replacing \(\angle DAB\) by equal angle \(\angle CDA\), we obtain \(\angle ABC+\angle CDA=180^\circ\). The sum of opposite angles is \(180^\circ\), hence quadrilateral \(ABCD\) is cyclic.

We have \(\angle A+\angle C=74^\circ+106^\circ=180^\circ\). In a convex quadrilateral, if the sum of opposite angles is \(180^\circ\), then its vertices lie on one circle. Therefore \(A,B,C,D\) lie on one circle.

In triangle \(ABC\), segment \(MN\) joins the midpoints of two sides, so \(MN\parallel AC\) and \(MN=\frac{1}{2}AC\). In triangle \(CDA\), segment \(PQ\) also joins the midpoints of two sides, so \(PQ\parallel AC\) and \(PQ=\frac{1}{2}AC\). Therefore \(MN\parallel PQ\) and \(MN=PQ\). If one pair of opposite sides in a quadrilateral is both equal and parallel, then it is a parallelogram. Hence \(MNPQ\) is a parallelogram.

By the midline theorem, \(MN=PQ=\frac{1}{2}AC=\frac{11}{2}\), and \(NP=QM=\frac{1}{2}BD=\frac{15}{2}\). Thus the perimeter is \(2\cdot\frac{11}{2}+2\cdot\frac{15}{2}=11+15=26\).

Triangles \(AOD\) and \(COB\) are similar: \(\angle AOD=\angle COB\) as vertical angles, and \(\angle ADO=\angle CBO\) because \(AD\parallel BC\). Therefore \(\frac{AO}{OC}=\frac{DO}{OB}=\frac{AD}{BC}=\frac{21}{14}=\frac{3}{2}\). Hence \(AO:OC=DO:OB=3:2\).

Since \(MN\parallel BC\), in triangle \(ABC\) the line through midpoint \(M\) of side \(AB\), parallel to \(BC\), meets \(AC\) at its midpoint. Hence \(P\) is the midpoint of \(AC\). Then \(MP\) is a midline of triangle \(ABC\), so \(MP=\frac{1}{2}BC\). Also \(PN\) is a midline of triangle \(ACD\), so \(PN=\frac{1}{2}AD\). Therefore \(MP:PN=BC:AD\).

By condition \(\angle BAC=\angle CAD\). Since \(AD\parallel BC\), \(\angle CAD=\angle ACB\). Therefore \(\angle BAC=\angle ACB\), so in triangle \(ABC\), sides \(AB\) and \(BC\) are equal. In a parallelogram, equality of adjacent sides means that all sides are equal. Hence \(ABCD\) is a rhombus.

From \(AO=OC\) and \(BO=OD\), the diagonals bisect each other. Hence \(ABCD\) is a parallelogram. A parallelogram with equal diagonals has all angles right: this follows by comparing triangles \(ABC\) and \(DCB\). Therefore \(ABCD\) is a rectangle.

Angle \(\angle CDE\) is formed by rays \(DC\) and \(DE\), and angle \(\angle CBE\) by rays \(BC\) and \(BE\). Both angles stand on the same segment \(CE\). If two points \(B\) and \(D\) see segment \(CE\) under equal angles, then they lie on one circle with \(C\) and \(E\). Therefore \(B,C,D,E\) lie on one circle.

In triangle \(ABC\), segment \(EP\parallel AB\), where \(E\) lies on \(BC\) and \(P\) lies on \(AC\). Therefore \(\triangle CPE\sim\triangle CAB\). From \(BE:EC=1:2\), we have \(CE:CB=2:3\). Hence \(CP:CA=2:3\). Thus \(AP:PC=(CA-CP):CP=1:2\).

Since \(M\) is the midpoint of \(AB\) and \(MN\parallel BC\), in triangle \(ABC\), point \(P\) is the midpoint of \(AC\), and \(MP=\frac{1}{2}BC\). Similarly, in triangle \(ABD\), the line through midpoint \(M\), parallel to \(AD\), meets \(BD\) at its midpoint \(Q\), and \(MQ=\frac{1}{2}AD\). Since \(AD>BC\), point \(Q\) lies farther from \(M\) than \(P\), so \(PQ=MQ-MP=\frac{AD}{2}-\frac{BC}{2}=\frac{AD-BC}{2}\).

Let \(AD=a\), \(BC=b\). From the similarity of triangles \(AOD\) and \(COB\), we get \(DO:OB=a:b\), so \(OB:DB=b:(a+b)\) and \(DO:DB=a:(a+b)\). In triangle \(BDA\), segment \(OX\parallel AD\), hence \(\frac{OX}{AD}=\frac{OB}{DB}=\frac{b}{a+b}\). Thus \(OX=\frac{ab}{a+b}\). In triangle \(DBC\), segment \(OY\parallel BC\), hence \(\frac{OY}{BC}=\frac{DO}{DB}=\frac{a}{a+b}\). Thus \(OY=\frac{ab}{a+b}\). Therefore \(OX=OY\). For \(a=12\), \(b=6\), we get \(OX=OY=\frac{12\cdot6}{18}=4\), so \(XY=8\).

In a parallelogram, \(\angle A=\angle C\). Since the quadrilateral is cyclic, opposite angles are supplementary, so \(\angle A+\angle C=180^\circ\). But these angles are equal, hence each is \(90^\circ\). Similarly, the remaining angles are also right. Therefore \(ABCD\) is a rectangle.

In an isosceles trapezoid, base angles are equal, so triangles \(DAB\) and \(ADC\) are congruent by two sides and the included angle: \(AD\) is common, \(AB=CD\), and the base angles at \(AD\) are equal. Therefore \(BD=AC\). Since the diagonals of a trapezoid are divided in the ratio of the bases, \(AO:OC=DO:OB=AD:BC=18:10=9:5\). The diagonals \(AC\) and \(BD\) are equal and divided in the same ratio \(9:5\), so their larger parts are equal: \(AO=DO\). Answer: \(AO:OC=9:5\).