Question

Ex 5 pls dau coroana plus 50 points.​

Answer 1

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Answer 2

$\it a)\\ \\ \left.\begin{aligned} \it \widehat{BOM}=\widehat{BOA}+\widehat{AOM}=90^o+\widehat{AOM}\\ \\ \it \widehat{BOM}=\widehat{POM}+\widehat{BOP}=90^o+\widehat{BOP} \end{aligned}\right\} \Rightarrow \widehat{AOM}=\widehat{BOP}\ \ \ \ \ (1)$

$\it [OA-bisectoare\ pentru\ \widehat{MON} \Rightarrow \widehat{MON}=2\cdot \widehat{AOM} \ \ \ \ \ (2)$

$\it [OB-bisectoare\ pentru\ \widehat{POQ} \Rightarrow \widehat{POQ}=2\cdot \widehat{BOP}\ \stackrel{(1)}{=}\ 2\cdot\widehat{AOM}\ \ \ \ \ (3)\\ \\ (2),\ (3) \Rightarrow \widehat{MON}=\widehat{POQ}$

$\it b)\\ \\ \widehat{NOQ}=\widehat{NOP}+\widehat{POQ}\ \stackrel{a)}{=}\ \widehat{NOP}+\widehat{MON}=\widehat{MOP}=90^o \Rightarrow ON\perp OQ$