Question

в треугольнике АBD угол ABD равен 90 градусов.Докажите что прямая BD является касательной к окружности с центром в точке А и радиусом равным АВ.пРОШУ РЕШИТЕ МНЕ НАДО ОЧЕЬН ЗАРАНЕЕ БЛАГОДАРЮ срочно

Answer 1

Если радиус= AB, то окружность касается с прямой BD в точке B. При этом, AB перпендикулярна BD. Значит радиусу перпендикулярен BD, приведенный в точку B. Из этого следует, что BD-касательная.
                                                                       Что и требовалось доказать

Answer 2

Proof that line BD is tangent to the circle with center A and radius AB

To prove that line BD is tangent to the circle with center A and radius AB, we can use the fact that angle ABD is 90 degrees.

Let's assume that line BD is not tangent to the circle. In that case, there would be a point C on the circle such that line BD intersects the circle at point C.

Now, let's consider triangle ABC. Since AB is a radius of the circle, it is equal to AC. Additionally, angle ABC is also 90 degrees because it is formed by a radius and a tangent line at point B.

Therefore, triangle ABC is an isosceles right triangle, with AB = AC and angle ABC = 90 degrees.

However, this contradicts the fact that angle ABD is also 90 degrees. In a triangle, the sum of all angles must be 180 degrees.

Since the assumption that line BD intersects the circle is false, we can conclude that line BD is indeed tangent to the circle with center A and radius AB.

Thus, we have proved that line BD is tangent to the circle.