Question

В треугольнике АВС АВ = 4 см. ВС=3 см, АС=5 см. Докажите, что АВ – отрезок касательной, проведенной из точки А к окружности с центром в точке С и радиусом, равным 3 см.P.S. Еще рисунок сделайте и прикрепить сюда, пожалуйста

Answer 1

А вот насчет как доказать

Answer 2

Proof that AB is the tangent to the circle with center C and radius 3 cm

To prove that AB is the tangent to the circle with center C and radius 3 cm, we need to show that AB is perpendicular to the radius drawn from the center of the circle to the point of tangency.

Given: - Triangle ABC with AB = 4 cm, AC = 5 cm, and BC = 3 cm.

To begin the proof, let's draw the triangle ABC and the circle with center C and radius 3 cm:


Now, let's assume that AB is not the tangent to the circle. This means that there is another line passing through point A that intersects the circle at a different point, let's call it D:


Since AD is not the tangent, it must intersect the circle at point D. Let's consider the line segment CD:


Now, we have two cases to consider:

Case 1: CD is shorter than AB:

In this case, we can see that triangle ACD is a right triangle, with AC as the hypotenuse. According to the Pythagorean theorem, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

Using the given lengths, we have:

AC^2 = AD^2 + CD^2

Substituting the values, we get:

5^2 = AD^2 + CD^2

25 = AD^2 + CD^2

Since AD and CD are both positive lengths, their squares are also positive. Therefore, the sum of their squares cannot be equal to 25. This contradicts the assumption that CD is shorter than AB.

Case 2: CD is longer than AB:

In this case, we can see that triangle BCD is a right triangle, with BC as the hypotenuse. Using the same logic as in Case 1, we can apply the Pythagorean theorem to triangle BCD:

BC^2 = BD^2 + CD^2

Substituting the values, we get:

3^2 = BD^2 + CD^2

9 = BD^2 + CD^2

Again, since BD and CD are both positive lengths, their squares are also positive. Therefore, the sum of their squares cannot be equal to 9. This contradicts the assumption that CD is longer than AB.

Since both cases lead to contradictions, our assumption that AB is not the tangent to the circle is false. Therefore, AB must be the tangent to the circle with center C and radius 3 cm.

Thus, we have proved that AB is the tangent to the circle with center C and radius 3 cm.

Conclusion

In conclusion, we have shown that AB is the tangent to the circle with center C and radius 3 cm. This was proven by assuming the contrary and showing that it leads to contradictions. Therefore, we can confidently state that AB is indeed the tangent to the circle.