Теорема о трех перпендикулярах с нуля || Геометрия 10 класс || Стереометрия

Link to the file with problems for independent solution: https://vk.com/wall-199104221_531 Problems solved: #1. Line MA is perpendicular to the plane of right triangle ABC (where ∠C = 90°). Prove that triangle MCB is a right triangle. #2. Line MC is perpendicular to the plane of rhombus ABCD, where O is the intersection point of its diagonals. a) Prove that MO ⊥ BD; b) Prove that MA ⊥ BD. #3. Segment AM is perpendicular to the plane of triangle ABC and has a length of 24 cm. Find the distance from point M to line BC if AB=AC=20 cm, BC=24 cm. Timecodes: 00:00 Oblique, projection, line 00:55 Three-perpendicular theorem 02:10 Problem 1 04:55 Problem 2 a) 05:49 Problem 2 b) 06:57 Problem 3 Notes: 1) When solving, it's better to write the full phrase "by the three-perpendicular theorem" instead of "by the 3⊥ theorem." In the video, this is done only for the sake of clarity. 2) In the video, I was careless; in problems, we usually talk about the perpendicularity of segments, not lines. 3) In the picture illustrating the theorem of three perpendiculars, it is better to depict the projection, perpendicular and inclined plane as straight lines rather than segments.