Each side of `DeltaABC` is the polar of the opposite vertex with respect to a circle with centre P. Then P is ortho centre of `DeltaABC`.

Answer

To prove that point P is the orthocenter of triangle ABC when each side of triangle ABC is the polar of the opposite vertex with respect to a circle with center P, we can follow these steps:

### Step 1: Understand the Definitions
- **Polar of a Point**: The polar of a point with respect to a circle is the locus of points where tangents drawn from the point to the circle are perpendicular to the radius at the point of tangency.
- **Orthocenter**: The orthocenter of a triangle is the point where the three altitudes intersect.

### Step 2: Identify the Polar Relationships
Let’s denote the sides of triangle ABC as follows:
- Side BC is the polar of vertex A.
- Side AC is the polar of vertex B.
- Side AB is the polar of vertex C.

### Step 3: Establish the Polar Condition
Since each side of triangle ABC is the polar of the opposite vertex with respect to the circle centered at P, we can write:
- The point A lies on the polar of A, which is line BC.
- The point B lies on the polar of B, which is line AC.
- The point C lies on the polar of C, which is line AB.

### Step 4: Use the Properties of Polars
From the properties of polars, we know that if a point lies on the polar of another point, the line connecting these two points is perpendicular to the line connecting the center of the circle to the point on the polar.

### Step 5: Establish Perpendicularity
1. Since A lies on the polar of A (line BC), the line AP is perpendicular to line BC.
2. Since B lies on the polar of B (line AC), the line BP is perpendicular to line AC.
3. Since C lies on the polar of C (line AB), the line CP is perpendicular to line AB.

### Step 6: Conclude the Position of P
Since P is the point from which the perpendiculars to the sides of triangle ABC are drawn, P must be the intersection point of the altitudes of triangle ABC. Therefore, P is the orthocenter of triangle ABC.

### Final Conclusion
Thus, we have shown that if each side of triangle ABC is the polar of the opposite vertex with respect to a circle centered at P, then P is indeed the orthocenter of triangle ABC.

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