Parallelogram

Draw the diagonal $AC.$ In the triangles $ABC$ and $CDA$ the side $AC$ is common, ${\angle BAC=\angle DCA}$ (alternate angles, ${AB\parallel DC},$ transversal $AC$) and ${\angle BCA=\angle DAC}$ (alternate angles, ${AD\parallel BC}$). By ASA the triangles are congruent, so ${AB=CD},$ ${BC=DA}$ and ${\angle B=\angle D}.$ The diagonal $BD$ gives ${\angle A=\angle C}$ in the same way.

The angles $A$ and $B$ are co-interior angles (${AD\parallel BC},$ transversal $AB$), so ${\alpha+\beta=180^\circ}.$

In the triangles $AOB$ and $COD$ we have ${AB=CD}$ and two pairs of equal alternate angles, ${\angle OAB=\angle OCD}$ and ${\angle OBA=\angle ODC}.$ By ASA they are congruent, so ${AO=OC}$ and ${BO=OD}.$

Opposite sides equal. If ${AB=CD}$ and ${BC=DA},$ the triangles $ABC$ and $CDA$ are congruent by SSS. Then ${\angle BAC=\angle DCA},$ and these are alternate angles, so ${AB\parallel DC};$ in the same way ${AD\parallel BC}.$

One pair parallel and equal. If ${AB\parallel DC}$ and ${AB=DC},$ the triangles $ABC$ and $CDA$ are congruent by SAS (${\angle BAC=\angle DCA}$ as alternate angles). Then ${\angle BCA=\angle DAC},$ so ${AD\parallel BC}.$

Diagonals bisect each other. If ${AO=OC}$ and ${BO=OD},$ the triangles $AOB$ and $COD$ are congruent by SAS (vertical angles at $O$), so ${AB=CD}$ and ${\angle OAB=\angle OCD},$ that is ${AB\parallel DC}.$ Now one pair of opposite sides is parallel and equal.

Opposite angles equal. If ${\angle A=\angle C=\alpha}$ and ${\angle B=\angle D=\beta},$ then ${2\alpha+2\beta=360^\circ},$ so ${\alpha+\beta=180^\circ}.$ The co-interior angles $A$ and $B$ add up to $180^\circ,$ so ${AD\parallel BC};$ in the same way ${AB\parallel DC}.$

Cut off the right triangle $AED$ and move it to the side $BC\text{:}$ the parallelogram becomes a rectangle with the sides $a$ and $h_a,$ so ${S=a\cdot h_a}.$ The height $h_a$ is the leg opposite $\alpha$ in the right triangle $AED$ with hypotenuse ${AD=b},$ so ${h_a=b\sin\alpha}$ and ${S=ab\sin\alpha}.$ In the same way $h_b$ is the leg opposite ${\angle C=\alpha}$ in the right triangle $DFC$ with hypotenuse ${DC=a},$ so ${h_b=a\sin\alpha}.$

The diagonal ${d_2=BD}$ lies opposite the angle $\alpha$ in the triangle $ABD,$ and ${d_1=AC}$ lies opposite the angle ${\beta=180^\circ-\alpha}$ in the triangle $ABC.$ The law of cosines gives the two lengths, because ${\cos\beta=-\cos\alpha}.$ Adding them gives ${d_1^2+d_2^2=2(a^2+b^2)}.$

The diagonals divide the parallelogram into four triangles with sides $\frac{d_1}{2}$ and $\frac{d_2}{2}$ and the angle $\varphi$ or ${180^\circ-\varphi}$ between them. Each has the area ${\frac12\cdot\frac{d_1}{2}\cdot\frac{d_2}{2}\sin\varphi},$ because ${\sin(180^\circ-\varphi)=\sin\varphi},$ and four of them give ${S=\frac12d_1d_2\sin\varphi}.$

Opposite angles are equal, so the two angles are adjacent: ${x+(x+40^\circ)=180^\circ},$ ${x=70^\circ}.$ The angles are $70^\circ,$ $110^\circ,$ $70^\circ$ and $110^\circ.$

${AB+AD=15},$ so ${AD=6}$ cm.

${S}={10\cdot 6\cdot\sin 150^\circ}={60\cdot\frac12}={30}$ cm² (${\sin 150^\circ=\sin 30^\circ=\frac12}$).

${\frac{48}{8}=6}$ cm to the 8 cm side and ${\frac{48}{12}=4}$ cm to the 12 cm side.

${S=\frac12\cdot 6\cdot 8\cdot\sin 30^\circ=12}$ cm².

${d^2}={2(25+49)-36}={112},$ so ${d=\sqrt{112}=4\sqrt7\approx 10.58}$ cm.

Half of the other diagonal is ${\sqrt{100-36}=8},$ so the diagonal is 16 cm and the area is ${\frac{12\cdot 16}{2}=96}$ cm².

${d=\sqrt{36+64}=10}$ cm; the diagonals bisect each other, so the distance is 5 cm.

${a=\frac{8}{\sqrt2}=4\sqrt2}$ cm and ${S=\frac{8^2}{2}=32}$ cm².

${\angle DMA=\angle MAB}$ (alternate angles, ${AB\parallel DC}$) and ${\angle MAB=\angle DAM}$ ($AM$ is the bisector), so the triangle $ADM$ is isosceles, ${DM=AD=8}$ cm and ${CM}={CD-DM}={12-8}={4}$ cm.