Given the rectangular ABCD, pass a as SA ⊥ plane AC, then a as AE ⊥ sb to sb to e, and e to ef ⊥ SC to F (1) Confirmation: AF ⊥ SC; (2) If the plane AEF intersects SD with G, it is proved that Ag ⊥ SD

Given the rectangular ABCD, pass a as SA ⊥ plane AC, then a as AE ⊥ sb to sb to e, and e to ef ⊥ SC to F (1) Confirmation: AF ⊥ SC; (2) If the plane AEF intersects SD with G, it is proved that Ag ⊥ SD

It is proved that: (1)

If the hypotenuse ab of right triangle ABC is in plane α, the projection of right vertex C in α is C1 (C1 does not belong to ab), then △ ABC1 is a. right triangle B. obtuse angle triangle C. acute angle triangle D

B

The base of p-abc is a right triangle with AC as the hypotenuse. The projection of vertex P on the base is exactly the outer center of triangle ABC, PA = AB = 1, BC = radical 2 Then the angle between Pb and the bottom is

? B = 90 degrees  the outer center of ABC is just on the midpoint of AC,  if the midpoint of AC is O, then the projection of vertex P on the bottom is point O  the angle between Pb and bottom is ? Pao ? AC = √ (AB ^ 2 + BC ^ 2) = √ 3 ? Ao = AC / 2 = √ 3 / 2 ? the projection of vertex P on the bottom surface is point O ? if Pb is projected on the bottom surface, then it is Bo ? AP = 1

The projection of hypotenuse ab of right triangle △ ABC in plane α is C ', If the hypotenuse ab of the right triangle △ ABC is in the plane α, and the projection of the right vertex C in α is C ', then △ ABC' is___ The answer is an obtuse triangle, but I think it's a right triangle

The answer is right, obtuse angle. If point C is in a, it must be a right angle. But if C is raised, the AB side remains unchanged, but AC 'and BC' are both shorter, so it is an obtuse angle

The triangle ABC is a right triangle, AB is a hypotenuse, and the three vertices are on the same side of plane A. their projections in a are respectively a'b'c '. If the triangle a'b'c' is a regular triangle, and Aa1 = 3, BB1 = 5, CC1 = 4, then the area of triangle a'b'c 'is

If we make plane ade ‖ plane a'b'c ', intersection BB' in D, and intersection CC 'in E, then BD = 5-3 = 2, CE = 4-3 = 1, then △ ade ≌ △ a'b'c', let the side length of equilateral triangle = a from ab ≌ = BC ≌ = = = > (a? + 2? = = = = > (a? + 1?)) + [a? + (2-1)?] = = > A = √ 2 ≌ s

If the hypotenuse ab of the right triangle ABC is in the plane g, the angles of AC and BC with G are 30 ° and 45 ° respectively, and CD is the height of the hypotenuse ab

Let C be the vertical plane g of CE, then ∠ CAE = 30 degrees, ∠ CBE = 45 degrees. Let CE = a in a right triangle CAE, AC = 2A in a right triangle CBE, BC = √ 2a in a right triangle ABC, ab = √ 6acd = AC * BC / AB = 2A / √ 3sincde = CE / CD = A / (2A / √ 3) = √ 3 / 2 angle CDE = 60 degrees

In the right triangle ABC, given B (- 3,0), C (3,0), then the trajectory equation of the right vertex A is What is the range of Y or x?

Let a (x, y)
AB^2=[x-(-3)]^2+(Y-0)^2
AC^2=(x-3)^2+(Y-0)^2
BC^2=(3+3)^2=36
Because AB ^ 2 + AC ^ 2 = BC ^ 2
So [x - (- 3)] ^ 2 + (y-0) ^ 2 + (x-3) ^ 2 + (y-0) ^ 2 = 36
So x ^ 2 + y ^ 2 = 9

Given the hypotenuse ab of right triangle ABC, and a (- 5,1), B (3-2), find the trajectory equation of vertex C

AB^2=(1+2)^2+(-5-3)^2=73
Let C coordinate be (x, y)
AC^2=(x+5)^2+(y-1)^2
BC^2=(x-3)^2+(y+2)^2
So there is: (x + 5) ^ 2 + (Y-1) ^ 2 + (x-3) ^ 2 + (y + 2) ^ 2 = 73
After simplification, the trajectory equation is obtained

If the vertex a (- 1,0) B (1,0) of the right triangle ABC, then the trajectory equation of the right vertex C is obtained

Let C (x, y), (x + 1) 2 + y 2 + (x-1) 2 + y 2 = 4
The results show that x 2 + y 2 = 1 (x ≠ ± 1, y ± 0)
The trajectory equation of right angle vertex C is: x 2 + y 2 = 1 (x ≠± 1, y ± 0)

Given the point B (- 2,1) and point C (3,2), if the right triangle ABC has BC as the hypotenuse, then the trajectory equation of the right vertex A is the best Given the point B (- 2,1) and point C (3,2), if the right triangle ABC takes BC as the hypotenuse, then the trajectory equation of the right vertex A is It's better to answer by voice and speak more carefully

Let a (x, y), by Pythagorean theorem: AB 2 + AC 2 = BC 2
Then: (x + 2) 2 + (Y-1) 2 + (x-3) 2 + (Y-2) 2 = (- 2-3) 2 + (1-2)?)
The trajectory equation of point a is obtained as follows: x 2 - x + y 2 - 3y-4 = 0