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7.3: Поперечна властивість
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$\measuredangle ABC + \measuredangle BCD \equiv 0$ або $\measuredangle ABC + \measuredangle BCD \equiv \pi.$ \(2 \cdot (\measuredangle ABC + \measuredangle BCD) \equiv 2 \cdot (\measuredangle ABC + ... $\measuredangle ABC + \measuredangle BCD \equiv 0$ або $\measuredangle ABC + \measuredangle BCD \equiv \pi.$ $2 \cdot (\measuredangle ABC + \measuredangle BCD) \equiv 2 \cdot (\measuredangle ABC + \measuredangle BCD') \equiv 0$ . Тому $2 \cdot \measuredangle BCD \equiv 2 \cdot BCD'$ і за допомогою вправи 2.4.2 $D' \in (CD)$ , або еквівалентно лінія $(CD)$ збігається з $(CD')$ .