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stas0107

Ответ:

Пошаговое объяснение:

PR||EF

Т.к. в ∆ PRE PE=PR, то <PER=<PRE,

а т.к. <PER=<REF, то <PRE=<REF.

<PRE и <REF-накрест лежащие, образованные прямыми РR и ЕF и секущей RE, они равны, следовательно, PR||EF

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