\( 2^{\circ} \) Show that \( A M N=A C B \) and that : \[ \widehat{A C B}+\widehat{N M B}=180^{\circ} \text {. } \] 10 Consider \( (C) \) a circle of center \( O \) and of radius \( R \). \( [O A] \) and \( [O B] \) are two perpendicular radii of \( (C) \). Consider \( C \) a point of the major arc \( \overparen{A B} \). The tangents at \( A \) and \( C \) to this circle intersect at \( M \). The line \( (C M) \) cuts the line \( (O B) \) at \( P \). Let \( D \) be the orthogonal projection of \( M \) on \( (O B) \). \( 1^{\circ} \) Show that the triangle \( P M O \) is isosceles of main vertex \( P \).
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Show that \(AMN = ACB\): Since \(AM\) and \(AC\) are tangents to the circle, we know that \(AM\) and \(AC\) are perpendicular to the radii \(OA\) and \(OC\) respectively. Therefore, \(AM\) is parallel to \(OC\) and \(AC\) is parallel to \(OM\). This means that Show more…
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