Question

Точка М не принадлежит плоскости ромба АВСД. Точка О-точка пересечения диагоналей ромба. Довести,что прямая МО перпендикулярна плоскости АВС,если МА=МС,МВ=МД.

Answer 1

АВСД -ромб.
Диагонали ромба перпендикулярны и точкой пересечения делятся пополам. В треугольнике ВМД ВМ=МД, ВО=ОД, МО - общая сторона.
Треугольники МОВ=МОД, треугольник ВМД равнобедренный. 
По тем же основаниям треугольник АМС равнобедренный.
МО является высотой обоих треугольников, перпендикулярна ВД и АС. 
Если прямая перпендикулярна двум пересекающимся прямым в плоскости, то она перпендикулярна этой плоскости. ⇒ 
МО⊥ плоскости ромба и плоскости треугольника АВС, принадлежащего этой плоскости, что и требовалось доказать. 

Answer 2

Introduction

In this problem, we are given a rhombus ABCD and a point M that does not belong to the plane of the rhombus. Point O is the intersection of the diagonals of the rhombus. We need to prove that the line MO is perpendicular to the plane of ABCD, given that MA = MC and MB = MD.

Proof

To prove that line MO is perpendicular to the plane of ABCD, we can use the properties of a rhombus and the concept of perpendicularity.

1. Properties of a Rhombus: - In a rhombus, the diagonals are perpendicular bisectors of each other. - The diagonals of a rhombus bisect each other at right angles. - The diagonals of a rhombus divide it into four congruent triangles.

2. Given Information: - Point M does not belong to the plane of the rhombus ABCD. - Point O is the intersection of the diagonals of the rhombus. - MA = MC and MB = MD.

3. Proof: - Since MA = MC and MB = MD, we can conclude that triangles MAB and MCD are congruent by the Side-Angle-Side (SAS) congruence criterion. - Therefore, angle MAB = angle MCD and angle MBA = angle MDC. - Since the diagonals of a rhombus bisect each other at right angles, we can conclude that angle MAB = angle MDC and angle MBA = angle MCD are right angles. - Thus, we have two pairs of opposite angles that are right angles in triangles MAB and MCD. - By the definition of a plane, any three non-collinear points determine a unique plane. - Therefore, the points A, B, and C lie on the same plane. - Since the diagonals of a rhombus divide it into four congruent triangles, we can conclude that triangles MAB and MCD lie on the same plane as ABCD. - Thus, the line MO, which passes through the intersection of the diagonals O, is perpendicular to the plane of ABCD.

Conclusion

In conclusion, we have proved that the line MO is perpendicular to the plane of the rhombus ABCD, given that MA = MC and MB = MD. This is based on the properties of a rhombus and the congruence of triangles MAB and MCD.