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ID number:921927
Published: 09.07.2007.
Language: Latvian
Level: College/University
Literature: n/a
References: Not used

Pieņemsim, ka plaknē novilkta taisne a – simetrijas ass, un atlikts punkts B, kurš nepieder taisnei. Lai atrastu kādam dotajam punktam atbilstošo punktu B1, novelkam perpendikulu no punkta B pret taisni a, kurš krusto to punktā O. Pagarināsim nogriezni BO otrā pus plaknē no taisnes a un uz tā atliksim nogriezni OB1 tā, ka OB=OB1. Punktu B1 sauc par punkta B aksiāli simetrisko punktu attiecībā pret simetrijas asi a vai īsāk – attiecībā pret taisni a. Tātad punkts B ir punkta B1 simetriskais punkts.

Definīcija: divus punktus B un B1 sauc par simetriskiem pret taisni a, ja nogrieznis BB1 ir perpendikulārs taisnei a, un taisne a iet caur nogriežņa BB1 viduspunktu. Taisni a sauc par simetrijas asi, bet punktus B un B1 – par aksiāli simetriskiem punktiem.…

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