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 \( P \). \( 2^{\circ} \) Show that \( O P M=2 \widehat{A O M} \). \( 3^{\circ} \) Show that the points \( O, D, C \) and \( M \) are on the same circle . \( 4^{\circ} \) Show that the quadrilateral \( O C D M \) is an isosceles trapezoid.

Let's analyze the given geometric configuration step by step to prove each part of the problem. ### **Given:** - Circle \( (C) \) with center \( O \) and radius \( R \). - Radii \( [OA] \) and \( [OB] \) are perpendicular. - \( C \) is a point on the major arc \( \overparen{AB} \). - Tangents at points \( A \) and \( C \) intersect at point \( M \). - Line \( CM \) intersects \( OB \) at point \( P \). - \( D \) is the orthogonal projection of \( M \) onto \( OB \). ### **Problem 1: Show that the triangle \( \triangle PMO \) is isosceles with main vertex \( P \).** **Proof:** 1. **Coordinate System Setup:** - Place \( O \) at the origin \((0, 0)\). - Let \( OA \) lie along the positive \( x \)-axis, so \( A = (R, 0) \). - Since \( [OA] \) and \( [OB] \) are perpendicular, \( B = (0, R) \). 2. **Coordinates of Point \( C \):** - Let \( C \) be at an angle \( \theta \) from \( OA \), where \( \theta > 90^\circ \) (since \( C \) is on the major arc \( \overparen{AB} \)). - Thus, \( C = (R \cos \theta, R \sin \theta) \). 3. **Equations of the Tangents:** - **Tangent at \( A \):** Since \( OA \) is along the \( x \)-axis, the tangent at \( A \) is vertical: \( x = R \). - **Tangent at \( C \):** The slope of \( OC \) is \( \tan \theta \), so the slope of the tangent at \( C \) is \( -\cot \theta \). - Equation: \( y - R \sin \theta = -\cot \theta (x - R \cos \theta) \). 4. **Finding Point \( M \):** - Intersection of the tangents: Set \( x = R \) in the tangent at \( C \): \[ y - R \sin \theta = -\cot \theta (R - R \cos \theta) \] \[ y = R \sin \theta + R \frac{\cos \theta (1 - \cos \theta)}{\sin \theta} = \frac{R (1 - \cos \theta)}{\sin \theta} \] - Therefore, \( M = \left(R, \frac{R (1 - \cos \theta)}{\sin \theta}\right) \). 5. **Finding Point \( P \):** - Line \( CM \) has slope \( -\cot \theta \). - Equation of \( CM \): \( y = -\cot \theta \cdot x + \frac{R}{\sin \theta} \). - Intersection with \( OB \) (\( x = 0 \)): \( P = \left(0, \frac{R}{\sin \theta}\right) \). 6. **Calculating Distances:** - \( PO = \frac{R}{\sin \theta} \). - \( PM \): \[ PM = \sqrt{R^2 + \left(\frac{R (1 - \cos \theta)}{\sin \theta} - \frac{R}{\sin \theta}\right)^2} = \sqrt{R^2 + \left(\frac{-R \cos \theta}{\sin \theta}\right)^2} = \sqrt{R^2 + \frac{R^2 \cos^2 \theta}{\sin^2 \theta}} = \frac{R}{\sin \theta} \] - \( MO \): \[ MO = \sqrt{R^2 + \left(\frac{R (1 - \cos \theta)}{\sin \theta}\right)^2} \] (Not directly required for this part) 7. **Conclusion:** - Since \( PO = PM \), triangle \( \triangle PMO \) is isosceles with \( P \) as the main vertex. ### **Answer to Part 1:** 1. **\( \triangle PMO \) is isosceles with \( P \) as the main vertex because \( PO = PM \).**