Kvadrāti \(ABCD\) un \(A_{1}B_{1}C_{1}D_{1}\) atrodas paralēlās plaknēs; abiem virsotnes uzrādītas pulksteņa rādītāja kustības virzienā. Pierādīt, ka \(AA_{1}^{2}+CC_{1}^{2}=BB_{1}^{2}+DD_{1}^{2}\).
Apzīmējam \(ABCD\) centru un malas garumu attiecīgi ar \(X\) un \(x,\ A_{1}B_{1}C_{1}D_{1}\) centru un malas garumu attiecīgi ar \(Y\) un \(y\), bet \(\overrightarrow{XY}=\vec{\omega}\). Tad
\[\begin{gathered} AA_{1}^{2}+CC_{1}^{2}=\left(\overrightarrow{AX}+\overrightarrow{\omega}+\overrightarrow{YA_{1}}\right)^{2}+\left(\overrightarrow{CX}+\overrightarrow{\omega}+\overrightarrow{YC_{1}}\right)^{2}= \\ =AX^{2}+YA_{1}^{2}+CX^{2}+YC_{1}^{2}+2 \omega^{2}+2 \overrightarrow{\omega}(\underbrace{\overrightarrow{AX}+\overrightarrow{CX}}_{\overrightarrow{0}}+\underbrace{\overrightarrow{YA_{1}}+\overrightarrow{YC_{1}}}_{\overrightarrow{0}})+2 \overrightarrow{AX} \cdot \overrightarrow{YA_{1}}+2 \overrightarrow{CX} \cdot \overrightarrow{YC_{1}}= \\ =x^{2}+y^{2}+2 \omega^{2}+2\left(\overrightarrow{AX} \cdot \overrightarrow{YA_{1}}+\overrightarrow{CX} \cdot \overrightarrow{YC_{1}}\right) \end{gathered}\]
Līdzīgi izsakot \(BB_{1}^{2}+DD_{1}^{2}\), iegūstam, ka jāpierāda vienādība\[\overrightarrow{AX} \cdot \overrightarrow{YA_{1}}+\overrightarrow{CX} \cdot \overrightarrow{YC_{1}}=\overrightarrow{BX} \cdot \overrightarrow{YB_{1}}+\overrightarrow{DX} \cdot \overrightarrow{YD_{1}}\]
Šīs vienādības pareizība seko no tā, ka \(|\overrightarrow{AX}|=|\overrightarrow{CX}|=|\overrightarrow{BX}|=|\overrightarrow{DX}|\), \(\left|\overrightarrow{YA_{1}}\right|=\left|\overrightarrow{YC_{1}}\right|=\left|\overrightarrow{YB_{1}}\right|=\left|\overrightarrow{YD_{1}}\right|\) un \(\sphericalangle\left(\overrightarrow{AX}, \overrightarrow{YA_{1}}\right)=\sphericalangle\left(\overrightarrow{CX}, \overrightarrow{YC_{1}}\right)=\sphericalangle\left(\overrightarrow{BX}, \overrightarrow{YB_{1}}\right)=\sphericalangle\left(\overrightarrow{DX}, \overrightarrow{YD_{1}}\right)\).