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.

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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 $$ PO = PM $$.**
$$ \triangle PMO $$ is isosceles with $$ P $$ as the main vertex. **Simplified Answer:** $$ \triangle PMO $$ is isosceles with $$ P $$ as the main vertex.

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