The user wants me to create an HTML output with two parts:
1. Analysis of the sample question
2. A list of 5 similar problems with solutions and answer keys.
The sample problem is about a pyramid S.ABCD with a rhombus base, SA perpendicular to the plane (ABCD). It asks which statement is false regarding angles between lines/planes and perpendicularity.
Let me analyze the problem type:
– Geometry of space (Hình học không gian) – Grade 12
– Topic: Relationships between lines and planes, angles between lines and planes, perpendicular relationships
– Specifically about pyramids with a perpendicular side edge to the base
– The notation [S;BD;C] likely means the angle between line BD and line SC? Or maybe it’s the angle between plane and line? Let me check:
From the solution: [S;BD;C] = (OS, OC) = ∠SOC. So [S;BD;C] seems to be the angle between line BD and line SC? Or perhaps it’s a notation for the angle between two lines where S is a point on one line and BD is the other line?
Actually, [S;BD;C] = (OS, OC) — the angle between OS and OC. So S and BD gives OS? That’s interesting. Let me think…
Actually, reading more carefully: [S;BD;C] is probably the angle between the line SC and the line BD? But they compute (OS, OC) — no wait, OS and OC are vectors from O to S and O to C. The angle between BD (or its projection) and SC.
Let me reconsider: In some Vietnamese geometry notation, [A;BC;D] means the angle between line AD and the plane (ABC)? Or maybe the angle between line AB and line CD? No…
From the sample: [S;BD;C] is computed as (OS, OC) = ∠SOC. So it seems they interpret [S;BD;C] as the angle between the line from S to BD (maybe the perpendicular from S to BD?) and the line from C to BD? No…
Actually, looking at the options:
A. [S;BD;C] = ∠SOA — but solution says it’s ∠SOC, so A is false (answer).
B. [S;AC;B] = 90° — angle between SB and AC? Or between S and AC relative to B?
C. [S;BD;A] = ∠SOA — angle between SA and BD?
D. (SAC) ⊥ (SBD) — planes are perpendicular.
Let me deduce the notation: [X;YZ;W] where X, W are points and YZ is a line segment. This might mean the angle between line XW and line YZ? Or maybe the angle between line through X perpendicular to YZ and line through W? Or the dihedral angle?
Wait, I think in Vietnamese geometry textbooks, [A;BC;D] or similar notation is sometimes used to denote the angle between two lines: specifically, the notation [S;BD;C] means the angle formed by the intersection of the plane containing S and BD and the plane containing C and BD? Or actually, I think [S;BD;C] represents the angle between the lines from S to BD and from C to BD? No…
Let me read the solution: “[S;BD;C] = (OS, OC) = ∠SOC”
So [S;BD;C] is computed as the angle between vectors OS and OC. So it seems [S;BD;C] means the angle at point O (or some intersection) formed by the two lines, where O is… O is the center of the rhombus.
Actually, looking at the geometry: S.ABCD has base rhombus ABCD, center O. SA ⟂ (ABCD). For option A, [S;BD;C]: BD is on the base, SC is a side edge. The angle between SC and BD — we project: from the solution, they compute angle between OS and OC.
Let me think again: In a rhombus, O is the center, AC ⟂ BD (since AC and BD are perpendicular diagonals in a rhombus? No, wait: in a rhombus, the diagonals are perpendicular only if it’s a square or a rhombus. Actually, in a rhombus, diagonals are perpendicular only in special cases like a square or a rhombus that is also a kite. In a general rhombus, diagonals are not necessarily perpendicular. But the problem says “hình thoi” which means rhombus, and its diagonals are perpendicular? Actually, yes, in a rhombus (hình thoi), the diagonals are perpendicular bisectors of each other. So AC ⟂ BD.
Wait, actually: In a rhombus, the diagonals are perpendicular to each other. Yes, that’s correct. So O is the intersection of AC and BD, with AC ⟂ BD.
Now, given SA ⟂ (ABCD), SA ⟂ AC and SA ⟂ BD.
For [S;BD;C], this might be the angle between the line through S and BD… Hmm.
Let me search my memory about Vietnamese geometry notation. I’ve seen something like: The notation [a;b] for angle between lines a and b. And [A;BC;D] might mean the angle between two lines: the line through A and the projection of A onto BC? Or the angle between line AD and line BC?
Actually, I think I recall: In some Vietnamese textbooks, the notation (A;BC;D) or [A;BC;D] means the angle between line AB and line CD? Or the angle between two lines, determined by three points where the middle one is on the intersection?
Wait, reading more carefully: “Khẳng định nào sau đây là SAI?” – which statement is FALSE.
Option A: [S;BD;C] = ∠SOA
Option B: [S;AC;B] = 90°
Option C: [S;BD;A] = ∠SOA
Option D: (SAC) ⊥ (SBD)
Solution: [S;BD;C] = (OS, OC) = ∠SOC. So A claims it’s ∠SOA, but it’s actually ∠SOC. So A is false.
Option C: [S;BD;A] = (OS, OA) = ∠SOA. So C is true.
Option B: [S;AC;B]: Should be 90°. Let’s check: [S;AC;B] might mean angle between SB and AC? Since SA ⟂ AC and … Actually, if [S;AC;B] is angle between SB and AC, then we need to check if SB ⟂ AC. In a rhombus, AC ⟂ BD, but does AC ⟂ SB? Not necessarily. Hmm…
Actually, wait: Maybe [X;YZ;W] means the angle between the line through X and YZ? Or maybe it’s the angle in the triangle formed by X, a point on YZ, and W?
Let me look at option B more carefully. [S;AC;B] = 90°. The solution didn’t evaluate this, but the answer is A. So B, C, D are all true.
Let me verify each:
D: (SAC) ⟂ (SBD): Since SA ⟂ (ABCD), SA ⟂ BD. Also, AC ⟂ BD (rhombus diagonals). So BD ⟂ (SAC). Since BD ⊂ (SBD), (SBD) ⟂ (SAC). So D is true.
C: [S;BD;A] = ∠SOA. We need to interpret [S;BD;A] as the angle between SA and BD? No — if it were angle between SA and BD, since SA ⟂ BD (because SA ⟂ (ABCD)), the angle would be 90°, not ∠SOA.
[S;BD;A] must be the angle between the planes (SBD) and (ABD)? Or between lines from S and A to BD?
From the solution to A, [S;BD;C] = (OS, OC). So it seems [X;YZ;W] = the angle between OX and OW where O is the intersection of the line from X and from W with YZ?
Actually, I think the notation [A;BC;D] might mean the angle between the plane (ABC) and the plane (DBC)? No…
Let me think of it differently. Maybe [A;BC;D] is the angle between the lines from A to BC and from D to BC. In other words, the angle between the two lines when you drop perpendiculars from A and D to line BC. That is, O is the foot of the perpendicular from A and D to BC? But in the solution, O is the center of the rhombus, not necessarily the foot of perpendicular from S and C to BD.
Wait, O is the center of the rhombus. In the solution for [S;BD;C], they use O as the intersection point. So O is on BD. And the angle is between OS and OC. So O is on BD.
So [S;BD;C] might mean: take the line BD, take points S and C on opposite sides, and draw lines from S and C to some point on BD? Or project S and C onto BD?
Let me check: Since O is the center of the rhombus and lies on BD. And OS is the line from O to S, OC is from O to C. So [S;BD;C] is the angle at O on BD formed by OS and OC. So essentially, it’s the angle between the lines from O to S and from O to C.
Similarly, for [S;BD;A], it would be angle between OS and OA (since A is on BD? No, A is not on BD… but OA is a line from O to A. So O is still on BD. The angle is between OS and OA.
For [S;AC;B], it would be the angle between OS and OB? Since O is on AC? Actually, O is the intersection of AC and BD. So for [S;AC;B], O would be on AC, and we compute angle between OS and OB. And the claim is 90°. Is that true? OS and OB: Are they perpendicular?
Let’s check: OS is from O to S, OB is from O to B. Since O is the center, OB ⟂ AC (in rhombus, diagonals are perpendicular). But OS is not necessarily perpendicular to OB. However, SA ⟂ (ABCD), and O is in the base. We have triangle SOB. Is ∠SOB = 90°? Not necessarily. So if [S;AC;B] means angle between OS and OB at O (on AC), then it’s not necessarily 90°. So the statement says it’s 90°, which should be false if it’s not 90°, but the answer is A is false, meaning B, C, D are true. So [S;AC;B] has to be 90°.
Let me recheck: In a rhombus, do diagonals bisect each other perpendicularly? Yes. So AC ⟂ BD at O.
Now, SA ⟂ (ABCD), so SA ⟂ AC and SA ⟂ BD. But does SB ⟂ AC? Or does OS ⟂ OB? Let’s compute.
In a rhombus, OB is along BD, and OC is along AC. OB ⟂ OC.
SA ⟂ (ABCD), so SA ⟂ OB and SA ⟂ OC.
For triangle SOB, if we consider OS and OB, is OS ⟂ OB?
– OB is in the base
– OS has components: OA (or O to A) and AS.
Actually, O is the center. OS = OA + AS (vector). OB is perpendicular to OA (since OA is along AC, OB is along BD, and AC ⟂ BD). But OS has an OA component and an AS component. OB ⟂ OA, and OB ⟂ AS (since SA ⟂ (ABCD) and OB is in the base), so OB ⟂ (OA + AS) = OS. So yes, OS ⟂ OB. Since OB ⟂ OA (diagonals perpendicular) and OB ⟂ SA (SA ⟂ base), so OB ⟂ plane (SOA)? Actually OB ⟂ OA and OB ⟂ SA, so OB ⟂ plane (SOA), so OB ⟂ OS. So OS ⟂ OB.
So angle between OS and OB is 90°, so [S;AC;B] = 90° is true.
Similarly, for [S;BD;A] = ∠SOA: O on BD, angle between OS and OA. Is this ∠SOA? Yes. And is this equal to ∠SOA? Well, it’s the angle between OS and OA in the triangle SOA. So yes.
For [S;BD;C] = ∠SOA: But the actual angle between OS and OC is ∠SOC, not ∠SOA. So A is false.
So the notation [X;YZ;W] means: The angle at the intersection point of line YZ with… Actually, O is the intersection point of line YZ (BD) with line XW? No, X=S and W=C, and YZ=BD. O is the intersection of BD with what? With AC?
Let me see: For [S;BD;C], O is on BD. But O is also the foot of the perpendicular from where? Actually O is the foot of the perpendicular from A and C onto BD? In a rhombus, the diagonals intersect at O, so O is the midpoint of both diagonals. So O is on BD, and also O is the midpoint.
So actually [S;BD;C] means: The angle between the lines SO and CO, where O is the foot of the perpendicular from S to BD? No, it can’t be because S is above the plane…
Actually, I think I understand now. The notation [A;BC;D] means the angle between the two planes (ABC) and (DBC) — that is, the dihedral angle along line BC. But then O doesn’t come into it.
Wait, the solution says [S;BD;C] = (OS, OC). If it’s the angle between planes (SBD) and (CBD), the common line is BD, and O would be a point on BD. The angle between the planes is the angle between two lines perpendicular to BD at O, one in each plane. In plane (SBD), the line through O perpendicular to BD is… Since BD ⟂ AC and BD ⟂ SA, BD ⟂ (SAC). So the line in (SBD) perpendicular to BD at O is the line from O to the intersection of (SBD) and (SAC)? Actually, (SBD) ∩ (SAC) = SO. And since BD ⟂ (SAC), BD ⟂ SO. So SO is perpendicular to BD. Similarly, in plane (CBD) (which is the base), the line through O perpendicular to BD is OC (since AC ⟂ BD). So the angle between the planes (SBD) and (CBD) is the angle between SO and OC, which is ∠SOC. So that makes sense: [S;BD;C] could mean the dihedral angle between the two half-planes (S, BD) and (C, BD) — the dihedral angle formed by the planes (SBD) and (CBD).
And similarly, [S;BD;A] means the dihedral angle between planes (SBD) and (ABD). The base plane (ABD) contains BD and A, and the perpendicular from O to BD in (ABD) is OA (since OA is along AC, and AC ⟂ BD). So the angle between (SBD) and (ABD) is the angle between SO and OA, which is ∠SOA. So C is correct.
For [S;AC;B], it means the dihedral angle between planes (SAC) and (BAC). Common line AC. In plane (SAC), the line through O perpendicular to AC: Since BD ⟂ AC and SA ⟂ AC? Wait, SA ⟂ (ABCD) so SA ⟂ AC. And OB is perpendicular to AC (since AC ⟂ BD). So in (SAC), the perpendicular to AC at O is? In (SAC), we have SA and SC. The line through O perpendicular to AC in plane (SAC) would be the line from O perpendicular to AC within (SAC). Since SA ⟂ AC, and SA is in (SAC), the perpendicular to AC in (SAC) through O would be parallel to SA. But O is not necessarily connected to S by a line perpendicular to AC at O. Actually, in (SAC), we have O on AC. The line OS: is OS perpendicular to AC? Let’s check: SA ⟂ AC, and OA is along AC. OS = OA + AS. OS · AC = (OA + AS) · AC = OA · AC + AS · AC. OA is along AC, so OA · AC = |OA||AC| ≠ 0. AS ⟂ AC, so AS · AC = 0. So OS is not perpendicular to AC generally. So OS is not the perpendicular line.
But we want the line in (SAC) through O perpendicular to AC. Since SA ⟂ AC, the line through O parallel to SA would be perpendicular to AC. But that’s not SO. So the perpendicular from O to AC in (SAC) is some line through O parallel to SA. Let’s call it L. And in (BAC) (the base), the perpendicular to AC at O is OB (since AC ⟂ BD and O is on BD). So the angle between these two perpendiculars is the angle between OB and the line through O parallel to SA. Since SA ⟂ (ABCD), and OB is in the base, SA ⟂ OB. So the angle is 90°. So the dihedral angle between (SAC) and (BAC) is 90°. So B is true.
So the interpretation: [A;BC;D] or [A;BC;D] means the dihedral angle between the two planes (ABC) and (DBC). In other words, the angle between the two half-planes that share the line BC, one containing A and one containing D.
So this notation is clear: [A;BC;D] = angle between planes (ABC) and (DBC). The line BC is the common line, and A and D are points in the respective planes.
Now I understand the problem type. It’s about identifying true/false statements about the angles between planes (or dihedral angles) in a pyramid with a perpendicular side edge to the base.
Now I need to create 5 similar problems with solutions.
Let me think about variations:
1. Change the shape of the base (square, rectangle, regular polygon, etc.)
2. Change the perpendicular condition (not SA ⟂ base, but maybe some other edge is perpendicular, or the pyramid is regular)
3. Change the statements to check about angles or perpendicular relationships
Let me create 5 problems:
Problem 1: Similar to sample but with a square base instead of rhombus, and ask which statement is false.
Problem 2: With a rectangular base, and different statements about angles.
Problem 3: With a regular triangular pyramid (tetrahedron), and statements about angles.
Problem 4: With base being an isosceles trapezoid, and some perpendicular condition.
Problem 5: With base being a regular hexagon, and SA ⟂ base.
Actually, let me think about more specific problems.
Problem 1: Cho hình chóp S.ABCD có đáy là hình vuông tâm O, SA ⟂ (ABCD). Mệnh đề nào sau đây đúng? (which is true)
A. [S;BC;D] = ∠SOA
B. [S;AB;C] = ∠SOB
C. (SAB) ⊥ (SAD)
D. (SAB) ⟂ (SBC)
Actually, this is too different. Let me make 5 problems that are similar in structure to the sample.
Let me write problems that ask “Which statement is false?” or “Which statement is true?” with 4 options involving dihedral angle notation [A;BC;D] and perpendicularity of planes.
Problem 1: Cho hình chóp S.ABC có đáy ABC là tam giác đều, SA ⟂ (ABC). Khẳng định nào sau đây là SAI?
A. [S;BC;A] = ∠SIA (với I là trung điểm BC)
B. [A;SB;C] = ∠AIC (với I là trung điểm SB? No…)
Hmm, I need to think about the notation carefully.
Let me use the same structure: pyramid with base being a polygon with center O, and SA ⟂ base. The dihedral angle notation [point; line; point] means the angle between the two planes sharing that line, each containing one of the two points.
For a pyramid S.ABCD with base rhombus, center O, SA ⟂ base.
Let me design 5 similar problems:
Problem 1: Cho hình chóp S.ABCD có đáy là hình vuông tâm O, cạnh a, SA ⟂ (ABCD) và SA = a. Khẳng định nào sau đây là SAI?
Or I can just vary the base shape and the statements.
Actually, the sample used a rhombus base. Let me vary:
– Square base
– Rectangle base
– Equilateral triangle base
– Regular pentagon base? Or maybe just a simple trapezoid?
– Parallelogram base
For each, I need to construct 4 statements about dihedral angles and check which is true/false.
Let me also consider: The notation [A;BC;D] is specific. I need to make sure the problems are correct.
Let me design 5 problems:
Problem 1: Pyramid S.ABCD with square base, SA ⟂ (ABCD). Which statement is false?
Problem 2: Pyramid S.ABCD with rectangle base (AB ≠ AD), SA ⟂ (ABCD). Which statement is true?
Problem 3: Pyramid S.ABC with equilateral triangle base, SA ⟂ (ABC). Which statement is false?
Problem 4: Pyramid S.ABCD with rhombus base (different properties from sample), but SA ⟂ base. Which statement is true?
Problem 5: Pyramid S.ABCD with trapezoid base, SA ⟂ base. Which statement is false?
Let me design each carefully with correct dihedral angles.
Actually, let me design 5 problems that are very similar to the sample, just with different bases or conditions, to create meaningful practice problems.
Let me start with Problem 1:
Câu 1. Cho hình chóp S.ABCD có đáy là hình vuông tâm O, SA ⟂ (ABCD). Khẳng định nào sau đây là SAI?
A. [S;BC;A] = 90° (angle between (SBC) and (ABC) — dihedral angle at BC)
B. [S;AB;D] = ∠SOB? Actually, for square, let me compute.
For a square ABCD with center O:
– AB ⟂ BC, AB ⟂ AD
– Diagonals AC ⟂ BD at O.
SA ⟂ (ABCD).
A: [S;BC;A] = angle between (SBC) and (ABC). Common line: BC. In (SBC), the line through O’ (some point on BC) perpendicular to BC. Let’s take I as midpoint of BC. In (SBC), we need a line through I perpendicular to BC. Since SB is not necessarily perpendicular to BC. But in (SBC), we can consider: BC ⟂ AB (in base), and SA ⟂ BC, so BC ⟂ (SAB). So BC ⟂ SB. So at I (midpoint of BC), the line from I to S (or to some point in (SBC)) is… Actually, BC ⟂ SB and BC ⟂ AB, so in (SBC), SB ⟂ BC, so the line through B perpendicular to BC is SB. At I (midpoint of BC), the line through I perpendicular to BC in (SBC) is parallel to SB. In (ABC), the line through I perpendicular to BC is AB (since AB ⟂ BC in a square). So the angle between (SBC) and (ABC) is the angle between SB and AB, which is ∠SBA. Is ∠SBA = 90°? No, ∠SBA is angle in triangle SBA, with SA ⟂ AB, so ∠SBA is not 90° unless SA=0. So it’s some acute angle. So A is likely false.
But I need to make sure only one is false. Let me check other options.
B: [S;BD;C] = ∠SOA? For square base, O is center of both diagonals. [S;BD;C] = angle between (SBD) and (CBD). Common line BD. In (SBD), the perpendicular to BD at O: Since BD ⟂ AC (diagonals in square are perpendicular) and SA ⟂ BD, so BD ⟂ (SAC). Thus in (SBD), the line through O perpendicular to BD is OS (since OS is intersection of (SBD) and (SAC), and BD ⟂ (SAC), so BD ⟂ OS). In (CBD) (= base), the perpendicular to BD at O is OC (since AC ⟂ BD). So the angle is between OS and OC = ∠SOC. So [S;BD;C] = ∠SOC, not ∠SOA. So B might be false too if it says ∠SOA.
Hmm, I need to design statements that are all true except one. Let me think more carefully.
Let me design Problem 1 with square base:
Pyramid S.ABCD with square base ABCD, center O, SA ⟂ (ABCD).
Which statement is FALSE?
A. [S;AC;B] = 90° — angle between (SAC) and (BAC). Common line AC. In (SAC), the perpendicular to AC at O: Since AC ⟂ BD and AC ⟂ SA? Actually SA ⟂ AC. So the line through O perpendicular to AC in (SAC) is the line through O parallel to SA. In (BAC) (=base), the perpendicular to AC at O is OB (since AC ⟂ BD). So angle between the two lines is angle between OB and the line parallel to SA. SA ⟂ OB, so the angle is 90°. So A is TRUE.
B. [S;AB;C] = ∠SOC? No, that doesn’t make sense. Let me think of a different approach.
Actually, let me design the problems more carefully. I’ll pick statements that I can verify easily.
Let me for Problem 1 make the base a square, and have 4 statements:
A. (SAB) ⟂ (SBC)
B. [S;BD;A] = ∠SOA
C. [S;AC;B] = 90°
D. (SAC) ⟂ (SBD)
For square base:
A: (SAB) ⟂ (SBC)? BC ⟂ AB (square), BC ⟂ SA (SA ⟂ base), so BC ⟂ (SAB). BC ⊂ (SBC), so (SAB) ⟂ (SBC). TRUE.
B: [S;BD;A] = angle between (SBD) and (ABD). Common line BD. Perpendicular to BD at O: in (SBD), OS; in (ABD), OA (since AC ⟂ BD). Angle = ∠SOA. TRUE.
C: [S;AC;B] = 90°. Common line AC. Perpendicular to AC at O: in (SAC), line through O parallel to SA (since SA ⟂ AC); in (BAC), OB (since AC ⟂ BD). Angle between OB and SA is 90°. TRUE.
D: (SAC) ⟂ (SBD)? BD ⟂ (SAC) as shown, BD ⊂ (SBD), so (SAC) ⟂ (SBD). TRUE.
So all are true! I need one false.
Let me try making A false:
A. [S;BC;A] = 90° — this is angle between (SBC) and (ABC). Common line BC. In (SBC), the perpendicular to BC at B is SB (since BC ⟂ AB, BC ⟂ SA, so BC ⟂ (SAB), so BC ⟂ SB). In (ABC), the perpendicular to BC at B is AB. So angle between SB and AB = ∠SBA. Is ∠SBA = 90°? No, triangle SBA has SA ⟂ AB, so ∠SBA is some acute angle (not 90°). So A is false. But wait, the perpendiculars must be taken at the same point on BC. At B, we have SB ⟂ BC and AB ⟂ BC. At the midpoint I of BC, we’d have lines parallel to SB and AB. So yes, the angle is ∠SBA ≠ 90°. So A is false.
So Problem 1: square base, A is false, B,C,D are true.
But I need to make sure about option C: [S;AC;B] = 90°. In square base, AC ⟂ BD, SA ⟂ AC. At O, in (SAC), the perpendicular to AC is a line through O parallel to SA. In (BAC), perpendicular to AC is OB. The angle between them is 90°. TRUE.
So A is the only false one.
But the answer should be A with the correct reasoning.
Ok, so Problem 1 is ready.
Problem 2: Let me use a rectangle base with SA ⟂ base. Rectangle ABCD (AB > AD). Which statement is true?
A. (SAC) ⟂ (SBD)
B. [S;CD;A] = 90°
C. [S;AB;C] = ∠S? something
D. [S;AD;B] = ∠S? something
For rectangle base:
Diagonals are not perpendicular (only in square). So AC is not ⟂ BD in general.
A: (SAC) ⟂ (SBD)? SA ⟂ BD (since SA ⟂ base), but AC is not ⟂ BD, so BD is not ⟂ (SAC). Need BD ⟂ (SAC) which requires BD ⟂ AC and BD ⟂ SA. SA ⟂ BD holds, but BD ⟂ AC doesn’t hold for rectangle. So (SAC) ⟂ (SBD) is false for rectangle (unless it’s a square).
So A would be false. But I want the true one to be… Actually, the problem asks “Khẳng định nào sau đây là ĐÚNG?” (which is TRUE). I can vary between asking for true or false.
Let me design Problem 2 as a “which is true” question with rectangle base.
Actually, let me think of 4 statements where only one is true.
For rectangle base: SA ⟂ (ABCD).
Option A: (SAB) ⟂ (SBC) — Check: BC ⟂ AB (rectangle), BC ⟂ SA, so BC ⟂ (SAB). BC ⊂ (SBC), so (SAB) ⟂ (SBC). TRUE. But this might be too easy, and SA ⟂ (SBC)? No, SA ⊥ AB, SA ⟂ AD, but is SA ⟂ BC? SA ⟂ base, so SA ⟂ BC. Yes.
So A is true for rectangle. But I need only one true.
B: (SAC) ⟂ (SBD) — not necessarily true, as discussed.
C: [S;AB;D] = ∠SOD? Let me compute [S;AB;D] = angle between (SAB) and (DAB). Common line AB. In (SAB), perpendicular to AB at some point on AB: since SA ⟂ AB, SA is the perpendicular. In (DAB) (=base), perpendicular to AB at A is AD (rectangle). So angle between SA and AD is 90° (since SA ⟂ AD)? Actually SA ⟂ base, so SA ⟂ AD. So angle = 90°. So [S;AB;D] = 90°. This could be a true statement.
D: [S;AD;B] = ∠S? Similar reasoning: angle between (SAD) and (BAD), common line AD. Perpendicular in (SAD): SA; in (BAD): AB. SA ⟂ AB, so = 90°.
So C and D are both true. So my design is problematic.
Let me Design Problem 2 differently. Maybe a triangular pyramid.
Problem 2: Cho hình chóp S.ABC có đáy là tam giác đều cạnh a, tâm O, SA ⟂ (ABC). Khẳng định nào sau đây là SAI?
For this:
A. [S;BC;A] = ∠SIA (với I là trung điểm BC) — angle between (SBC) and (ABC). Common line BC. In (SBC), the line through I perpendicular to BC: Since BC ⟂ AI (equilateral triangle, AI median ⟂ BC), and BC ⟂ SA, so BC ⟂ (SAI). So in (SBC), SI ⟂ BC? Check: Since BC ⟂ (SAI), BC ⟂ SI. So SI ⟂ BC. In (ABC), AI ⟂ BC. So the angle is between SI and AI = ∠SIA. So A is TRUE.
B. [S;AB;C] = ∠SIC? No, let me think.
[S;AB;C] = angle between (SAB) and (CAB). Common line AB. I is midpoint of AB? In (SAB), we need a line through some point on AB perpendicular to AB. Let J be midpoint of AB. In (SAB), is SJ ⟂ AB? Not necessarily. Let’s check: AB ⟂ CJ (since CJ is median in equilateral triangle, but CJ ⟂ AB if triangle is equilateral? Yes, in equilateral triangle, all medians are altitudes. So CJ ⟂ AB. And SA ⟂ AB, so AB ⟂ (SCJ)? Actually AB ⟂ CJ and AB ⟂ SA, so AB ⟂ (SAC)? Wait, SA and CJ are not necessarily in the same plane. SA ⟂ AB and CJ ⟂ AB, but CJ is in base, SA is vertical, so AB ⟂ SA and AB ⟂ CJ, so AB ⟂ the plane through SA and CJ which is (SAC)? Actually SA and CJ intersect at A? No, CJ intersects AB at J, SA is at A. So the plane containing SA and CJ is not defined uniquely. But AB ⟂ SA and AB ⟂ CJ, so AB ⟂ the plane containing SA and CJ only if SA and CJ are in the same plane. They intersect at A and J? No, A is on AB, J is on AB. SA is at A, CJ is at J. They are not in the same plane unless A=J, which is not. So AB ⟂ (SAC) is not guaranteed. Let me think differently.
Actually, in (SAB), is there a line through J perpendicular to AB? In (SAB), we have SA ⟂ AB at A, and maybe SJ is not ⟂ AB. But we can take the perpendicular through J as a line parallel to SA through J. Since SA ⟂ AB, the line through J parallel to SA is perpendicular to AB and lies in (SAB) (since SA is in (SAB) and J is on AB in (SAB), the line through J parallel to SA is in (SAB)). So the perpendicular to AB in (SAB) at J is the line through J parallel to SA. In (CAB) (=base), the perpendicular to AB at J is CJ (since CJ ⟂ AB in equilateral triangle). So the angle between (SAB) and (CAB) is the angle between CJ and the line parallel to SA at J, which is 90° (since SA ⟂ (ABC), so SA ⟂ CJ). So [S;AB;C] = 90°. That could be a statement.
Hmm, let me use a
