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Tags: Geometri , R1

Vi kan vise at  \(\triangle PA_1B_2\) er formlik med \(\triangle PA_2B_1\).

Last ned Geogebra fil.

P er felles vinkel, og \(\angle A_2B_2A_1\) spenner over samme vinkelbuen som \( \angle A_1B_2A_2\). Vi kjenner to vinkler i begge trekantene og de er derfor formlike.

Dette betyr at \(\frac{PA_1}{PB_2}=\frac{PA_2}{PB_1}\)

Vi har altså at punktets potens er produktet

\(PA_1\cdot PB_1=PA_2\cdot PB_2\)

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