Question

В окружности проведены диаметры ab и cd. Докажите что хорды ac и bd параллельны.Помогите пожалуйста решить.

Answer 1

АО = ОВ, СО = OD как радиусы,

∠АОС = ∠BOD как вертикальные, ⇒

ΔАОС = ΔBOD по двум сторонам и углу между ними.

Значит ∠САО = ∠DBO, а эти углы - накрест лежащие при пересечении прямых АС и BD секущей АВ, значит

АС ║ BD.

Answer 2

Proof of the Parallel Chords in a Circle

To prove that the chords AC and BD are parallel, we can use the property that if a pair of opposite angles formed by the intersection of two lines are equal, then the lines are parallel. Let's proceed with the proof.

Proof:

Given: In a circle, diameters AB and CD are drawn.

We know that a diameter of a circle subtends a right angle to any point on the circle. Therefore, ∠AOC and ∠BOD are right angles.

From the properties of a circle, we also know that the angle subtended by a diameter at any point on the circle is a right angle. Therefore, ∠AOC = ∠BOD = 90°.

Now, let's consider the triangles AOC and BOD. - AO = OB (radii of the same circle) - CO = OD (radii of the same circle) - AOC = BOD = 90° (as discussed above)

By the Side-Angle-Side (SAS) congruence criterion, we can conclude that triangle AOC is congruent to triangle BOD.

As a result, corresponding sides of congruent triangles are equal. Therefore, AC = BD.

Since AC = BD, and they are opposite sides of the intersection of the diameters, we can conclude that AC and BD are parallel.

Hence, it is proved that the chords AC and BD are parallel.

I hope this helps! If you have further questions or need additional assistance, feel free to ask.