4-мысал. Кез келген үшбұрыштың орта сызығы табанына параллель және оның жартысына тең болатынын дәлелдеңдер (7-сызба).
![](data:image/png;base64,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) Шешуi. Үшбұрыштың бүйір қабырғаларының орталарын қосатын кесiндiнi, ол үшбұрыштың орта сызығы дейтiндiктен, EF орта сызығы болса, онда: болады.
Ал, Сөйтiп,
Бұдан векторлардың коллинеар болу анықтамасы бойынша
Болады және векторлар тең болса, ұзындықтары да тең болатындықтан: Сонымен, , болады екен.
Достарыңызбен бөлісу: |