Question

В треугольнике АBC угол А = 90 градусов. АН - высота треугольника. Вне плоскости АBC выбрана точка D, причем DB перпендикулярно BC и DB перпендикулярно AB Докажите, что прямая АН перпендикулярна к плоскости DBC

Answer 1

Если  DB перпендикулярна АВ и ВС, то она перпендикулярна самой плоскости АВС (по признаку перпендикулярности прямых и плоскостей).
Если АВС перпендикулярна DB, то она перпендикурна всей плоскости BDC по тому же признаку. 
То есть плоскость AВС перпендикулярна плоскости BDC.
Если BDC перпендикулярна АВС, то ИВС перпендикулярна прямой АН, лежащей в плоскости АВС по признаку перпендикулярности прямой и плоскости.

Answer 2

Given Information:

In triangle ABC, angle A is 90 degrees. AN is the height of the triangle. Point D is chosen outside the plane of ABC, such that DB is perpendicular to BC and DB is perpendicular to AB. We need to prove that line AN is perpendicular to the plane of DBC.

Proof:

To prove that line AN is perpendicular to the plane of DBC, we can show that line AN is perpendicular to any two lines in the plane of DBC.

Let's consider line AB and line BC.

1. Line AB is perpendicular to DB (given). 2. Line AB is perpendicular to AN (height of the triangle). 3. Therefore, line AB is perpendicular to the plane of DBC.

Next,

4. Line BC is perpendicular to DB (given). 5. Line BC is perpendicular to AN (height of the triangle). 6. Therefore, line BC is perpendicular to the plane of DBC.

Since line AN is perpendicular to both line AB and line BC, it is also perpendicular to the plane of DBC.

Therefore, we have proved that line AN is perpendicular to the plane of DBC.

Note: The proof assumes that the given information is correct and the lines AB, BC, and DB are in the same plane.