Question

Через точку C окружности с центром O проведена касательная AB, причем AC=CB. Докажите, что AO=OB. Желательно подробно и с рисуком. но подробно может и хватит. заранее спасибо

Answer 1

Треугольники АСО и СОВ равны по двум катетам
Катет СО - общий
АС=СВ
Из равенства треугольников АО=ОВ

Answer 2

Proof that AO = OB

To prove that AO = OB, we will use the properties of a tangent to a circle.

Let's consider a circle with center O and a tangent line AB, where AC = CB.

Step 1: Draw a diagram to visualize the problem.


Step 2: Since AC = CB, we can conclude that triangle ABC is an isosceles triangle.

Step 3: In an isosceles triangle, the angles opposite the equal sides are also equal.

Therefore, angle AOC = angle BOC.

Step 4: The sum of the angles in a triangle is 180 degrees.

So, angle AOC + angle BOC + angle AOB = 180 degrees.

Since angle AOC = angle BOC, we can rewrite the equation as:

2 * angle AOC + angle AOB = 180 degrees.

Step 5: In a circle, the angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the circumference.

Therefore, angle AOC = 1/2 * angle AOB.

Substituting this into the equation from step 4, we get:

2 * (1/2 * angle AOB) + angle AOB = 180 degrees.

Simplifying the equation, we have:

angle AOB + angle AOB = 180 degrees.

2 * angle AOB = 180 degrees.

Step 6: Divide both sides of the equation by 2:

angle AOB = 90 degrees.

Step 7: In a circle, any angle subtended by a diameter is a right angle.

Since angle AOB is subtended by the diameter AB, it is a right angle.

Step 8: In a right-angled triangle, the hypotenuse is twice the length of the shorter side.

Therefore, AO = OB.

Hence, we have proved that AO = OB.

I hope this explanation was helpful! Let me know if you have any further questions.