September 2018 1 11 Report
W trojkacie ABC, |AB|=, |BC|=7. Punkt D dzieli bok AB tak, ze |AD|:|DB|=3:2. Przez punkt D poprowadzono prosta rownolegla do boku AC przecinajaca bok BC w punkcie E. Oblicz: |AD|, |DB|, |BE|, |CE|..

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