Через конец а отрезка ав проведена плоскость а. через точку в и точку с, лежащую на отрезке ав, проведены параллельные прямые, пересекающие плоскость а в точках в1 и с1. докажите, что точки а, в1 и с1 лежат на одной прямой.

Прямые вв1 исс1 - параллельны, по определению параллельных прямых, они лежат в одной плоскости  β,  которая пересекает плоскость α.  если две плоскости  имеют общую точку,  то они имеют общую прямую, на которой лежат все их общие точки.  точки а, в1, с1 - общие точки плоскостей α  и β, т.е. они лежат на одной прямой.   точка а лежит на прямой  ав,  по условию , значит она лежит в плоскости β.  если две точки прямой лежат в плоскости, то и все точки этой прямой лежат в данной плоскости.

По свойству биссектрис трапеции они образовывают при боковых сторонах равнобедренные треугольники. Тогда ВК = АВ = 25 см, СК = СД = 30 см, тогда ВС = ВК + СК = 25 + 30 = 55 см.

Построим высоты ВН и СМ. Четырехугольник НВСМ прямоугольник, тогда НМ = ВС = 55 см.

В прямоугольном треугольнике СДМ определим длину катета ДМ.

ДМ2 = СД2 – СМ2 = 900 – 576 = 324.

ДМ = 18 см.

В прямоугольном треугольнике АВН определим длину катета АН.

АН2 = АВ2 – ВН2 = 625 – 576 = 49.

ДМ = 7 см.

Тогда АД = АН + НМ + ДМ = 7 + 55 + 18 = 80 см.

Определим площадь трапеции.

Sавсд = (ВС + АД) * ВН / 2 = (55 + 80) * 24 / 2 = 1620 см2.

ответ: Площадь трапеции равна 1620 см2.

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Объяснение:

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