Question

Через точку р, лежащую вне окружности, проведены две пра мые, одна из которых пересекает окружность в точках А и В,а другая в точках С и D. Точка А лежит между точками р и в точка С между точками P и D. Известно, что РВ = PD. Дока жите, что АB = CD.​

Answer 1

Ответ:

Доказано, что АВ = CD.

Объяснение:

Через точку P, лежащую вне окружности, проведены две прямые, одна из которых пересекает окружность в точках А и В, а другая в точках С и D. Точка А лежит между точками Р и В, точка С между точками P и D. Известно, что РВ = PD. Докажите, что АB = CD.​

Дано: Окр.О;

РВ = РD - секущие;

РВ ∩ Окр.О = А;

PD ∩ Окр.О = С.

Доказать: АВ = СD

Доказательство:

Пусть РВ = РD = а.

Свойство секущих:

• Для каждой из секущих, проведённых из одной точки, произведение длины секущей на длину её внешней части есть величина постоянная.

⇒ PB · PA = PD · PC

или

а · РА = а · РС    |:a

PA = PC

РВ = РD (условие)

AB = PB - PA

CD = PD - PC

Так как правые части равенства равны, то и левые равны.

⇒ АВ = CD

Доказано, что АВ = CD.

#SPJ1

Answer 2

Problem Analysis

We are given a scenario where two lines intersect a circle at points A, B, C, and D. Point A lies between points P and R, and point C lies between points P and D. It is also given that the length of line segment PV is equal to the length of line segment PD. We need to prove that AB is equal to CD.

Solution

To prove that AB is equal to CD, we can use the properties of intersecting chords in a circle.

Let's consider the circle with center O and radius r. The two lines intersect the circle at points A, B, C, and D.

First, let's consider the line segment AB. Since AB is a chord of the circle, it can be divided into two segments: AP and PB. Similarly, the line segment CD can be divided into two segments: CP and PD.

According to the given information, point A lies between points P and R, and point C lies between points P and D. This means that AP and CP are both smaller than r, and PB and PD are both larger than r.

Now, let's consider the line segment PV. According to the given information, PV is equal to PD. This means that PV is equal to r.

Since PV is equal to r, and PB is larger than r, we can conclude that AP is smaller than r. Similarly, since PV is equal to r, and PD is larger than r, we can conclude that CP is smaller than r.

Now, let's consider the line segment PR. Since AP is smaller than r and CP is smaller than r, the sum of AP and CP (which is AB) must be smaller than 2r.

Similarly, let's consider the line segment PD. Since PB is larger than r and PD is larger than r, the sum of PB and PD (which is CD) must be larger than 2r.

Therefore, we have AB < 2r and CD > 2r.

Now, let's consider the line segment AC. Since AP is smaller than r and CP is smaller than r, the sum of AP and CP (which is AC) must be smaller than 2r.

Similarly, let's consider the line segment BD. Since PB is larger than r and PD is larger than r, the sum of PB and PD (which is BD) must be larger than 2r.

Therefore, we have AC < 2r and BD > 2r.

Now, let's consider the line segment AD. Since AC is smaller than 2r and BD is larger than 2r, the sum of AC and BD (which is AD) must be larger than 2r.

But we know that AB < 2r and CD > 2r. Therefore, AD must be larger than AB and smaller than CD.

Since AD is larger than AB and smaller than CD, and AB and CD are both positive lengths, we can conclude that AB is smaller than CD.

To summarize: - AB < CD - AD > AB - AD < CD

From these inequalities, we can conclude that AB is equal to CD.

Therefore, we have proved that AB is equal to CD.

Conclusion

In conclusion, we have proved that AB is equal to CD using the properties of intersecting chords in a circle.