Two parallel planes What are the conditions and properties in spatial geometry? Are there ways to prove 2 planes are parallel? Let’s find out the details of the lecture about two parallel planes with TRƯỜNG THPT TRẦN HƯNG ĐẠO in the article below.
Relative position of two distinct planes
Given the line d and the plane (P). There are three cases:
(P), (Q) have 1 thing in common: (P) (Q) = a
(P), (Q) have nothing in common: (P) // (Q)
The concept of two parallel planes
Two planes are said to be parallel if they have no common ground.
Theorem 1
If the plane (α) contains two intersecting lines a, b and a, b are parallel to the plane (β), then (α) is parallel to (β).
+) To prove that two planes are parallel, we must prove that there are two intersecting lines belonging to one plane that are parallel to the other plane respectively.
+) To prove the line a // (Q), we prove the line a lies in the plane (P) // (Q).
Theorem 2
Through a point outside a given plane there is one and only one plane parallel to the given plane.
Consequent:
+) If the line d is parallel to the plane (α), then in (α) there is a line parallel to d and through d there is only one plane parallel to (α). Therefore, the line d is parallel to (α) we have to prove that d belongs to the plane (β) and has (α) // (β) ⇒ d // (α).
+) Two distinct planes parallel to the third plane are parallel to each other.
+) Let A point not lying on the plane (α). Every line that passes through A and is parallel to (α) lies in the plane through A and is parallel to (α).
Theorem 3
Given two parallel planes. If a plane intersects this plane, it also intersects the other plane and the two intersections are parallel.
Consequent: Two parallel planes intercept on two parallel lines of equal line segments.
Theorem 4: Thales .’s theorem
Three parallel double planes intercept on any two tangent lines proportionally corresponding line segments.
Exercises to apply two parallel planes
Question 1. Let S.ABCD pyramid with base ABCD a trapezoid where AD // BC and AD = 2BC. Let M and N be the midpoints of SA and AD respectively. Proof: (BMN) // (SCD).
The answer
Since N is the midpoint of AD, then:
The quadrilateral NBCD has: ND = BC and ND // BC so NDBC is a parallelogram
Infer: NB // CD NB // (SCD)
Triangle SAD has M, N being the midpoints of AS and AD . respectively
So: MN is the moving average of ∆ADS.
Infer: MN // SD MN // (SCD)
Are from
(BMN) // (SCD)
Verse 2. Let the pyramid S.ABCD have base as a parallelogram with center O. Let M, N, P be the midpoints SA, SB, SD and K, I are the midpoints of BC, OM, respectively.
a) Prove: (OMN) // (SCD)
b) Prove: (PMN) // (ABCD)
c) Proof: KI // (SCD)
The answer
a) We have: O, M are the midpoints of AC and SA, respectively, so OM // SC
Infer: OM // (SCD)
Analogue: ON // (SCD)
Then: (OMN) // (SCD)
b) We have: N, M are the midpoints of SB and SA, respectively, so MN // AB
Infer: MN // (ABCD)
Similar: PM // (ABCD)
So: (PMN) // (ABCD)
c) We have: O, K are the midpoints of AC and BC, respectively, so OK // AB
Inferred: OK // MN
Then: 5 points M, N, K, O, I are coplanar
From the above sentence (OMN) // (SCD) then KI // (SCD).
Verse 3. Let S.ABCD pyramid whose base is a parallelogram with center O. Let M, N be the midpoints SA, SD, respectively.
a) Prove: (OMN) // (SBC)
b) Let P, Q, R be the midpoints of AB, ON, SB respectively. Proof: PQ // (SBC) and (ROM) //
(SCD).
The answer
a) We have: O, M are the midpoints of AC and SA, respectively, so OM // SC
Inferred: OM // (SBC)
Analogue: ON // (SBC)
Then: (OMN) // (SBC)
b) We have: O, P are the midpoints of AC and BA respectively, so OP // CB
Infer: OP // (SBC) or P (OMN)
Otherwise: Q (OMN)
According to above (OMN) // (SBC) then PQ // (SBC).
We have: R, O are the midpoints of SB and BD, respectively, so RO // SD
Infer: RO // (SCD)
According to on OM // SC so OM // (SCD)
So: (ROM) // (SCD).
Verse 4. Let two parallelograms ABCD and ABEF share side AB and are not coplanar. Let I, J, K be the midpoints of AB, CD, and EF respectively. Prove:
a) (ADF) // (BCE)
b) (DIK) // (JBE)
The answer
a) We have: AD // BC
Infer: AD // (BCE)
Analogue: AF // (BCE)
Then: (ADF) // (BCE)
b) In parallelogram ABCD, I and J are the midpoints of AB and CD, respectively, so BI = DJ
Hence: IBJD is a parallelogram
Infer: DI // BJ should DI // (JBE)
In parallelogram ABEF, I, K are the midpoints of AB and EF, respectively, so IK // EF
Inferred: IK // (JBE)
So: (DIK) // (JBE)
Question 5. Let two parallelograms ABCD and ABEF lie on two different planes. On diagonals AC, BF takes points M, N such that MC = 2AM, NF = 2BN. Through M, N draw lines parallel to side AB, respectively, cut edges AD, AF in order at MfirstfemalefirstProve that:
a) MN // DE
b) USfirstWOMENfirst // (DEF)
c) (MNMfirstWOMENfirst) // (DEF)
The answer
a) Let I be the intersection of BM with CD.
Then we have:
On the other hand:
Then: MN // IF
Follow above
Candlestick
Infer: DI = CD = AB
Again: DI // EF
Hence: DEFI is a parallelogram or FI // DE
So: MN // DE
b) By the assumption that MMfirst // NNfirst (because they are parallel to AB) so M, MfirstN, FEMALEfirst flat copper.
Again: MMfirst // (DEF) (because of MMfirst // CD // AB)
And according to the above sentence, MN // DE should MN // (DEF)
So: (MMfirstWOMENfirstN) // (DEF)
Derived from: MOTHERfirstWOMENfirst // (DEF)
c) Proven in sentence b.
Verse 6. Let two parallelograms ABCD and ABEF lie on two distinct planes. Let I, J, K in order be the centroids of triangles ADF, ADC, BCE. Prove that: (IJK) // (CDFE).
The answer
We have: CD // EF // AB so CD and EF are coplanar.
Let M, N be the midpoints of DF and CD, respectively.
Then, since I, J are the centroids of triangles ADF and ADC, respectively, so:
⇒ IJ // MN IJ // (CDEF)
Otherwise, let P be the midpoint CE.
Then:
We have: ABCD, ABEF are parallelograms, so CDEF is also a parallelogram.
Then for M, P is the midpoint of two opposite sides of parallelogram CDEF
Should: MP // CD //AB
Infer: IK // MP // AB
Hence: ABPK is also a parallelogram
We have:
Infer: IK // MN
Then: IK // (CDEF)
So: (IJK) // (CDEF)
Verse 7. Let S.ABCD pyramid with base is a parallelogram with center O. Let M, N, P be the midpoints SA, BC, CD respectively.
a) Find the intersection of two planes (SAD) and (MOP).
b) Let E be the midpoint of SC and I the point on the side SA satisfying AI = 3IS. Find K = IE ∩ (ABC) and H = AB ∩ (EIN). Calculate the ratio
.
c) Let G be the centroid of triangle SBC. Find the cross-section of the pyramidal S.ABC cut by (IMG).
The answer
a) We have: O, P are the midpoints of AC and CD, respectively, so OP // AD
Inferred: OP // (SAD)
Then: the intersection of (SAD) and (OMP) is a line through M and is parallel to AD and intersects SD at the midpoint Q of SD.
b) Consider the plane (SAC) with:
Inference: IE cuts AC at K
Then: K = IE (ABC)
Applying Menelaus’ theorem in triangle SAC with three collinear points K, E, I has:
Applying Menelaus’ theorem in triangle ABC with three collinear points K, N, H has:
Then:
c) We see, the plane (IMG) is also the plane (SAG).
Since: G is the centroid of triangle SBC and N is the midpoint of BC
So: (IMG) (SBC) = SN
So: The cross-section of the pyramid S.ABC cut by (IMG) is triangle SAN
Verse 8. Let the pyramid S.ABCD have base ABCD as a parallelogram with center O. Let M, N be the midpoints of SA and CD, respectively. Let E be the intersection of AD and (BMN), I be the midpoint of ME and G = AN ∩ BD.
a) Find the point E and the intersection F of SD with the plane (BMN). Prove: FS = 2FD.
b) Prove: FG // (SAB) and (CDI) // (SAB).
c) Let H be the intersection of MN and SG. Proof: OH // GF.
The answer
a) In the plane (ABCD) stretched, BN intersects the line AD at E.
Then: E is the intersection of (BMN) with AD
Let F be the intersection of ME with SD.
Then: F is the intersection of SD with (BMN)
Because:
So: D is mid point of segment AE
From this, it follows that: SD and EM are the medians of the triangle SAE .
Infer: F is the centroid of triangle SAE
So: FS = 2FD
b) Triangle DGN and triangle BGA are similar, so
Thence inferred:
Should: FG // SB FG // (SAB)
We have: CD // AB and DI // MA
From that, infer: (CDI) // (SAB)
c) We have: G is the centroid of triangle ACD.
Applying Menelaus’ theorem to the triangle SAG with three collinear points M, H, N we have:
We also have:
According to the above proof, we also have: GF // SB
So: OH // GF
Verse 9. Let the pyramid S.ABCD have base ABCD as a parallelogram with center O. Let M be the midpoint of SC, N be the point on diagonal BD such that BD = 3BN.
a) Determine the intersection of the plane (SCD) and (SAB) and find T = DM ∩ (SAB). Calculate
.
MS:
b) Call K = AN ∩ Prove that: MK // (SBD).
c) Let I = AN DC, L = IM Calculate the ratio
and .
MS:
The answer
a) The planes (SAB) and (SCD) contain two parallel lines, AB and CD, respectively, so their intersection is the line d through S and d // AB // CD.
In the plane (SCD) stretched DM intersects d at T.
Then: T d T ∈ (SAB)
So: T = DM ∩ (SAB)
Because CD // DT, two triangles MCD and MST are similar.
Therefore:
So:
b) Because:
Therefore: N is the centroid of triangle ABC
It follows that K is the midpoint of BC .
Which leads to MK being the midline of the triangle SBC.
Should: MK // SB MK // (SBD)
c) Triangle IKC and triangle IAD are similar, so
Applying Menelaus’ theorem to triangle SCD with three points I, M, L collinear we have:
Applying Menelaus’ theorem to triangle IDL with three collinear points S, M, C we have:
Thence inferred:
Let h, k be the lengths of the altitudes of the triangles IKM and IAL drawn from M and L, respectively.
It is easy to see that:
So:
Verse 10. Given two squares ABCD and ABEF in two distinct planes. On the diagonals AC and BF, take the points M and N, respectively, such that AM = BN. Lines parallel to AB drawn from M, N intersect AD and AF at M’ and N’, respectively. Prove that:
a) (ADF) // (BCE)
b) (CDF) // (MM’N’N)
The answer
a) We have:
b) We have:
(first)
We also have:
(2)
And from the assumption we have:
(3)
From (3) infer: M’N’ // DF.
We also have: MM’ // NN’ // DC // FE
So: (CDF) // (MM’N’N)
Verse 11. Let ABC.A’B’C’ prism. Let I, J, K be the centroids of triangles ABC, ACC’, A’B’C’, respectively. Prove that: (IJK) // (BCC’B’) and (A’JK) // (AIB’).
The answer
+) Let M, N, P be the midpoints of BC, CC’ and B’C’, respectively.
According to the properties of the centroid of the triangle, we have:
Quadrilateral AMP’A’ is a parallelogram and has:
So: (IJK) // (BCC’B’)
+) Note that: plane (AIB’) is plane (AMB’), plane (A’JK) is plane (A’CP).
Because: AM // A’P, MB’ // CP (because quadrilateral B’MCP is a parallelogram)
So we have: (A’JK) // (AIB’).
Verse 12. Let the pyramid S.ABCD have base ABCD as trapezoid with big bottom AD and AD = 2BC, M ∈ BC. Let (P) be the plane passing through M, (P) // CD, (P) // SC, (P) intersects AD, SA, SB at N, P, Q, respectively.
a) Prove that: NQ // (SCD) and NP // SD.
b) Let H, K be the midpoints of SD and AD respectively. Prove that: (CHK) // (SAB).
The answer
a) From M we draw a line parallel to CD that cuts AD at N and cuts AB at E.
From M draw a line parallel to SC that intersects SB at Q.
Extend the SA cut EQ at P.
According to the construction, we infer: (EPN) // (SCD) and NQ ⊂ (EPN)
So: NQ // (SCD)
Because (P) // (SCD) and these two planes intersect (SAD) along the intersections NP and SD.
So we infer: NP // SD
b) We have: HK is the median of triangle SAD
Should: HK // SA (1)
Since K is the midpoint of AD, AK = BC.
Therefore: quadrilateral ABCK is a parallelogram
Infer: CK // AB (2)
From (1) and (2) infer: (CKH // (SAB).
Verse 13. Let the pyramid S.ABC have G as the centroid of triangle ABC. On segment SA, take two points M and N such that SM = MN = NA.
a) Prove that: GM // (SBC).
b) Let D be a point symmetrical to A through G. Prove that: (MCD) // (NBG).
c) Let H = DM ∩ (SBC). Prove that: H is the centroid of triangle SBC.
The answer
a) Let E be the mid point of BC. Then we have:
So: GM // (SBC)
b) From the assumption we deduce G, N are the midpoints of AD and AM respectively.
Hence: NG // MD (1)
Quadrilateral BDCG has E as the midpoint of the two diagonals, so it is a parallelogram.
Derived: BG // CD (2)
From (1) and (2) infer: (MCD) // (NBG)
c) We have: AE is the median of triangle SBC (3)
Applying Menelaus’ theorem to triangle SAE with three collinear points M, H, D we have:
(4)
From (3) and (4) infer: H is the centroid of triangle SBC.
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Two parallel planes What are the conditions and properties in spatial geometry? Are there ways to prove 2 planes are parallel? Let's find out the details of the lecture about two parallel planes with TRƯỜNG THPT TRẦN HƯNG ĐẠO in the article below.
Relative position of two distinct planes
Given the line d and the plane (P). There are three cases:
(P), (Q) have 1 thing in common: (P) (Q) = a
(P), (Q) have nothing in common: (P) // (Q)
The concept of two parallel planes
Two planes are said to be parallel if they have no common ground.
Theorem 1
If the plane (α) contains two intersecting lines a, b and a, b are parallel to the plane (β), then (α) is parallel to (β).
+) To prove that two planes are parallel, we must prove that there are two intersecting lines belonging to one plane that are parallel to the other plane respectively.
+) To prove the line a // (Q), we prove the line a lies in the plane (P) // (Q).
Theorem 2
Through a point outside a given plane there is one and only one plane parallel to the given plane.
Consequent:
+) If the line d is parallel to the plane (α), then in (α) there is a line parallel to d and through d there is only one plane parallel to (α). Therefore, the line d is parallel to (α) we have to prove that d belongs to the plane (β) and has (α) // (β) ⇒ d // (α).
+) Two distinct planes parallel to the third plane are parallel to each other.
+) Let A point not lying on the plane (α). Every line that passes through A and is parallel to (α) lies in the plane through A and is parallel to (α).
Theorem 3
Given two parallel planes. If a plane intersects this plane, it also intersects the other plane and the two intersections are parallel.
Consequent: Two parallel planes intercept on two parallel lines of equal line segments.
Theorem 4: Thales .'s theorem
Three parallel double planes intercept on any two tangent lines proportionally corresponding line segments.
Exercises to apply two parallel planes
Question 1. Let S.ABCD pyramid with base ABCD a trapezoid where AD // BC and AD = 2BC. Let M and N be the midpoints of SA and AD respectively. Proof: (BMN) // (SCD).
The answer
Since N is the midpoint of AD, then:
The quadrilateral NBCD has: ND = BC and ND // BC so NDBC is a parallelogram
Infer: NB // CD NB // (SCD)
Triangle SAD has M, N being the midpoints of AS and AD . respectively
So: MN is the moving average of ∆ADS.
Infer: MN // SD MN // (SCD)
Are from
(BMN) // (SCD)
Verse 2. Let the pyramid S.ABCD have base as a parallelogram with center O. Let M, N, P be the midpoints SA, SB, SD and K, I are the midpoints of BC, OM, respectively.
a) Prove: (OMN) // (SCD)
b) Prove: (PMN) // (ABCD)
c) Proof: KI // (SCD)
The answer
a) We have: O, M are the midpoints of AC and SA, respectively, so OM // SC
Infer: OM // (SCD)
Analogue: ON // (SCD)
Then: (OMN) // (SCD)
b) We have: N, M are the midpoints of SB and SA, respectively, so MN // AB
Infer: MN // (ABCD)
Similar: PM // (ABCD)
So: (PMN) // (ABCD)
c) We have: O, K are the midpoints of AC and BC, respectively, so OK // AB
Inferred: OK // MN
Then: 5 points M, N, K, O, I are coplanar
From the above sentence (OMN) // (SCD) then KI // (SCD).
Verse 3. Let S.ABCD pyramid whose base is a parallelogram with center O. Let M, N be the midpoints SA, SD, respectively.
a) Prove: (OMN) // (SBC)
b) Let P, Q, R be the midpoints of AB, ON, SB respectively. Proof: PQ // (SBC) and (ROM) //
(SCD).
The answer
a) We have: O, M are the midpoints of AC and SA, respectively, so OM // SC
Inferred: OM // (SBC)
Analogue: ON // (SBC)
Then: (OMN) // (SBC)
b) We have: O, P are the midpoints of AC and BA respectively, so OP // CB
Infer: OP // (SBC) or P (OMN)
Otherwise: Q (OMN)
According to above (OMN) // (SBC) then PQ // (SBC).
We have: R, O are the midpoints of SB and BD, respectively, so RO // SD
Infer: RO // (SCD)
According to on OM // SC so OM // (SCD)
So: (ROM) // (SCD).
Verse 4. Let two parallelograms ABCD and ABEF share side AB and are not coplanar. Let I, J, K be the midpoints of AB, CD, and EF respectively. Prove:
a) (ADF) // (BCE)
b) (DIK) // (JBE)
The answer
a) We have: AD // BC
Infer: AD // (BCE)
Analogue: AF // (BCE)
Then: (ADF) // (BCE)
b) In parallelogram ABCD, I and J are the midpoints of AB and CD, respectively, so BI = DJ
Hence: IBJD is a parallelogram
Infer: DI // BJ should DI // (JBE)
In parallelogram ABEF, I, K are the midpoints of AB and EF, respectively, so IK // EF
Inferred: IK // (JBE)
So: (DIK) // (JBE)
Question 5. Let two parallelograms ABCD and ABEF lie on two different planes. On diagonals AC, BF takes points M, N such that MC = 2AM, NF = 2BN. Through M, N draw lines parallel to side AB, respectively, cut edges AD, AF in order at MfirstfemalefirstProve that:
a) MN // DE
b) USfirstWOMENfirst // (DEF)
c) (MNMfirstWOMENfirst) // (DEF)
The answer
a) Let I be the intersection of BM with CD.
Then we have:
On the other hand:
Then: MN // IF
Follow above
Candlestick
Infer: DI = CD = AB
Again: DI // EF
Hence: DEFI is a parallelogram or FI // DE
So: MN // DE
b) By the assumption that MMfirst // NNfirst (because they are parallel to AB) so M, MfirstN, FEMALEfirst flat copper.
Again: MMfirst // (DEF) (because of MMfirst // CD // AB)
And according to the above sentence, MN // DE should MN // (DEF)
So: (MMfirstWOMENfirstN) // (DEF)
Derived from: MOTHERfirstWOMENfirst // (DEF)
c) Proven in sentence b.
Verse 6. Let two parallelograms ABCD and ABEF lie on two distinct planes. Let I, J, K in order be the centroids of triangles ADF, ADC, BCE. Prove that: (IJK) // (CDFE).
The answer
We have: CD // EF // AB so CD and EF are coplanar.
Let M, N be the midpoints of DF and CD, respectively.
Then, since I, J are the centroids of triangles ADF and ADC, respectively, so:
⇒ IJ // MN IJ // (CDEF)
Otherwise, let P be the midpoint CE.
Then:
We have: ABCD, ABEF are parallelograms, so CDEF is also a parallelogram.
Then for M, P is the midpoint of two opposite sides of parallelogram CDEF
Should: MP // CD //AB
Infer: IK // MP // AB
Hence: ABPK is also a parallelogram
We have:
Infer: IK // MN
Then: IK // (CDEF)
So: (IJK) // (CDEF)
Verse 7. Let S.ABCD pyramid with base is a parallelogram with center O. Let M, N, P be the midpoints SA, BC, CD respectively.
a) Find the intersection of two planes (SAD) and (MOP).
b) Let E be the midpoint of SC and I the point on the side SA satisfying AI = 3IS. Find K = IE ∩ (ABC) and H = AB ∩ (EIN). Calculate the ratio
.
c) Let G be the centroid of triangle SBC. Find the cross-section of the pyramidal S.ABC cut by (IMG).
The answer
a) We have: O, P are the midpoints of AC and CD, respectively, so OP // AD
Inferred: OP // (SAD)
Then: the intersection of (SAD) and (OMP) is a line through M and is parallel to AD and intersects SD at the midpoint Q of SD.
b) Consider the plane (SAC) with:
Inference: IE cuts AC at K
Then: K = IE (ABC)
Applying Menelaus' theorem in triangle SAC with three collinear points K, E, I has:
Applying Menelaus' theorem in triangle ABC with three collinear points K, N, H has:
Then:
c) We see, the plane (IMG) is also the plane (SAG).
Since: G is the centroid of triangle SBC and N is the midpoint of BC
So: (IMG) (SBC) = SN
So: The cross-section of the pyramid S.ABC cut by (IMG) is triangle SAN
Verse 8. Let the pyramid S.ABCD have base ABCD as a parallelogram with center O. Let M, N be the midpoints of SA and CD, respectively. Let E be the intersection of AD and (BMN), I be the midpoint of ME and G = AN ∩ BD.
a) Find the point E and the intersection F of SD with the plane (BMN). Prove: FS = 2FD.
b) Prove: FG // (SAB) and (CDI) // (SAB).
c) Let H be the intersection of MN and SG. Proof: OH // GF.
The answer
a) In the plane (ABCD) stretched, BN intersects the line AD at E.
Then: E is the intersection of (BMN) with AD
Let F be the intersection of ME with SD.
Then: F is the intersection of SD with (BMN)
Because:
So: D is mid point of segment AE
From this, it follows that: SD and EM are the medians of the triangle SAE .
Infer: F is the centroid of triangle SAE
So: FS = 2FD
b) Triangle DGN and triangle BGA are similar, so
Thence inferred:
Should: FG // SB FG // (SAB)
We have: CD // AB and DI // MA
From that, infer: (CDI) // (SAB)
c) We have: G is the centroid of triangle ACD.
Applying Menelaus' theorem to the triangle SAG with three collinear points M, H, N we have:
We also have:
According to the above proof, we also have: GF // SB
So: OH // GF
Verse 9. Let the pyramid S.ABCD have base ABCD as a parallelogram with center O. Let M be the midpoint of SC, N be the point on diagonal BD such that BD = 3BN.
a) Determine the intersection of the plane (SCD) and (SAB) and find T = DM ∩ (SAB). Calculate
.
MS:
b) Call K = AN ∩ Prove that: MK // (SBD).
c) Let I = AN DC, L = IM Calculate the ratio
and .
MS:
The answer
a) The planes (SAB) and (SCD) contain two parallel lines, AB and CD, respectively, so their intersection is the line d through S and d // AB // CD.
In the plane (SCD) stretched DM intersects d at T.
Then: T d T ∈ (SAB)
So: T = DM ∩ (SAB)
Because CD // DT, two triangles MCD and MST are similar.
Therefore:
So:
b) Because:
Therefore: N is the centroid of triangle ABC
It follows that K is the midpoint of BC .
Which leads to MK being the midline of the triangle SBC.
Should: MK // SB MK // (SBD)
c) Triangle IKC and triangle IAD are similar, so
Applying Menelaus' theorem to triangle SCD with three points I, M, L collinear we have:
Applying Menelaus' theorem to triangle IDL with three collinear points S, M, C we have:
Thence inferred:
Let h, k be the lengths of the altitudes of the triangles IKM and IAL drawn from M and L, respectively.
It is easy to see that:
So:
Verse 10. Given two squares ABCD and ABEF in two distinct planes. On the diagonals AC and BF, take the points M and N, respectively, such that AM = BN. Lines parallel to AB drawn from M, N intersect AD and AF at M' and N', respectively. Prove that:
a) (ADF) // (BCE)
b) (CDF) // (MM'N'N)
The answer
a) We have:
b) We have:
(first)
We also have:
(2)
And from the assumption we have:
(3)
From (3) infer: M'N' // DF.
We also have: MM' // NN' // DC // FE
So: (CDF) // (MM'N'N)
Verse 11. Let ABC.A'B'C' prism. Let I, J, K be the centroids of triangles ABC, ACC', A'B'C', respectively. Prove that: (IJK) // (BCC'B') and (A'JK) // (AIB').
The answer
+) Let M, N, P be the midpoints of BC, CC' and B'C', respectively.
According to the properties of the centroid of the triangle, we have:
Quadrilateral AMP'A' is a parallelogram and has:
So: (IJK) // (BCC'B')
+) Note that: plane (AIB') is plane (AMB'), plane (A'JK) is plane (A'CP).
Because: AM // A'P, MB' // CP (because quadrilateral B'MCP is a parallelogram)
So we have: (A'JK) // (AIB').
Verse 12. Let the pyramid S.ABCD have base ABCD as trapezoid with big bottom AD and AD = 2BC, M ∈ BC. Let (P) be the plane passing through M, (P) // CD, (P) // SC, (P) intersects AD, SA, SB at N, P, Q, respectively.
a) Prove that: NQ // (SCD) and NP // SD.
b) Let H, K be the midpoints of SD and AD respectively. Prove that: (CHK) // (SAB).
The answer
a) From M we draw a line parallel to CD that cuts AD at N and cuts AB at E.
From M draw a line parallel to SC that intersects SB at Q.
Extend the SA cut EQ at P.
According to the construction, we infer: (EPN) // (SCD) and NQ ⊂ (EPN)
So: NQ // (SCD)
Because (P) // (SCD) and these two planes intersect (SAD) along the intersections NP and SD.
So we infer: NP // SD
b) We have: HK is the median of triangle SAD
Should: HK // SA (1)
Since K is the midpoint of AD, AK = BC.
Therefore: quadrilateral ABCK is a parallelogram
Infer: CK // AB (2)
From (1) and (2) infer: (CKH // (SAB).
Verse 13. Let the pyramid S.ABC have G as the centroid of triangle ABC. On segment SA, take two points M and N such that SM = MN = NA.
a) Prove that: GM // (SBC).
b) Let D be a point symmetrical to A through G. Prove that: (MCD) // (NBG).
c) Let H = DM ∩ (SBC). Prove that: H is the centroid of triangle SBC.
The answer
a) Let E be the mid point of BC. Then we have:
So: GM // (SBC)
b) From the assumption we deduce G, N are the midpoints of AD and AM respectively.
Hence: NG // MD (1)
Quadrilateral BDCG has E as the midpoint of the two diagonals, so it is a parallelogram.
Derived: BG // CD (2)
From (1) and (2) infer: (MCD) // (NBG)
c) We have: AE is the median of triangle SBC (3)
Applying Menelaus' theorem to triangle SAE with three collinear points M, H, D we have:
(4)
From (3) and (4) infer: H is the centroid of triangle SBC.
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Relative position of two distinct planes
Given the line d and the plane (P). There are three cases:
(P), (Q) have 1 thing in common: (P) (Q) = a
(P), (Q) have nothing in common: (P) // (Q)
The concept of two parallel planes
Two planes are said to be parallel if they have no common ground.
Theorem 1
If the plane (α) contains two intersecting lines a, b and a, b are parallel to the plane (β), then (α) is parallel to (β).
+) To prove that two planes are parallel, we must prove that there are two intersecting lines belonging to one plane that are parallel to the other plane respectively.
+) To prove the line a // (Q), we prove the line a lies in the plane (P) // (Q).
Theorem 2
Through a point outside a given plane there is one and only one plane parallel to the given plane.
Consequent:
+) If the line d is parallel to the plane (α), then in (α) there is a line parallel to d and through d there is only one plane parallel to (α). Therefore, the line d is parallel to (α) we have to prove that d belongs to the plane (β) and has (α) // (β) ⇒ d // (α).
+) Two distinct planes parallel to the third plane are parallel to each other.
+) Let A point not lying on the plane (α). Every line that passes through A and is parallel to (α) lies in the plane through A and is parallel to (α).
Theorem 3
Given two parallel planes. If a plane intersects this plane, it also intersects the other plane and the two intersections are parallel.
Consequent: Two parallel planes intercept on two parallel lines of equal line segments.
Theorem 4: Thales .’s theorem
Three parallel double planes intercept on any two tangent lines proportionally corresponding line segments.
Exercises to apply two parallel planes
Question 1. Let S.ABCD pyramid with base ABCD a trapezoid where AD // BC and AD = 2BC. Let M and N be the midpoints of SA and AD respectively. Proof: (BMN) // (SCD).
The answer
Since N is the midpoint of AD, then:
The quadrilateral NBCD has: ND = BC and ND // BC so NDBC is a parallelogram
Infer: NB // CD NB // (SCD)
Triangle SAD has M, N being the midpoints of AS and AD . respectively
So: MN is the moving average of ∆ADS.
Infer: MN // SD MN // (SCD)
Are from
(BMN) // (SCD)
Verse 2. Let the pyramid S.ABCD have base as a parallelogram with center O. Let M, N, P be the midpoints SA, SB, SD and K, I are the midpoints of BC, OM, respectively.
a) Prove: (OMN) // (SCD)
b) Prove: (PMN) // (ABCD)
c) Proof: KI // (SCD)
The answer
a) We have: O, M are the midpoints of AC and SA, respectively, so OM // SC
Infer: OM // (SCD)
Analogue: ON // (SCD)
Then: (OMN) // (SCD)
b) We have: N, M are the midpoints of SB and SA, respectively, so MN // AB
Infer: MN // (ABCD)
Similar: PM // (ABCD)
So: (PMN) // (ABCD)
c) We have: O, K are the midpoints of AC and BC, respectively, so OK // AB
Inferred: OK // MN
Then: 5 points M, N, K, O, I are coplanar
From the above sentence (OMN) // (SCD) then KI // (SCD).
Verse 3. Let S.ABCD pyramid whose base is a parallelogram with center O. Let M, N be the midpoints SA, SD, respectively.
a) Prove: (OMN) // (SBC)
b) Let P, Q, R be the midpoints of AB, ON, SB respectively. Proof: PQ // (SBC) and (ROM) //
(SCD).
The answer
a) We have: O, M are the midpoints of AC and SA, respectively, so OM // SC
Inferred: OM // (SBC)
Analogue: ON // (SBC)
Then: (OMN) // (SBC)
b) We have: O, P are the midpoints of AC and BA respectively, so OP // CB
Infer: OP // (SBC) or P (OMN)
Otherwise: Q (OMN)
According to above (OMN) // (SBC) then PQ // (SBC).
We have: R, O are the midpoints of SB and BD, respectively, so RO // SD
Infer: RO // (SCD)
According to on OM // SC so OM // (SCD)
So: (ROM) // (SCD).
Verse 4. Let two parallelograms ABCD and ABEF share side AB and are not coplanar. Let I, J, K be the midpoints of AB, CD, and EF respectively. Prove:
a) (ADF) // (BCE)
b) (DIK) // (JBE)
The answer
a) We have: AD // BC
Infer: AD // (BCE)
Analogue: AF // (BCE)
Then: (ADF) // (BCE)
b) In parallelogram ABCD, I and J are the midpoints of AB and CD, respectively, so BI = DJ
Hence: IBJD is a parallelogram
Infer: DI // BJ should DI // (JBE)
In parallelogram ABEF, I, K are the midpoints of AB and EF, respectively, so IK // EF
Inferred: IK // (JBE)
So: (DIK) // (JBE)
Question 5. Let two parallelograms ABCD and ABEF lie on two different planes. On diagonals AC, BF takes points M, N such that MC = 2AM, NF = 2BN. Through M, N draw lines parallel to side AB, respectively, cut edges AD, AF in order at MfirstfemalefirstProve that:
a) MN // DE
b) USfirstWOMENfirst // (DEF)
c) (MNMfirstWOMENfirst) // (DEF)
The answer
a) Let I be the intersection of BM with CD.
Then we have:
On the other hand:
Then: MN // IF
Follow above
Candlestick
Infer: DI = CD = AB
Again: DI // EF
Hence: DEFI is a parallelogram or FI // DE
So: MN // DE
b) By the assumption that MMfirst // NNfirst (because they are parallel to AB) so M, MfirstN, FEMALEfirst flat copper.
Again: MMfirst // (DEF) (because of MMfirst // CD // AB)
And according to the above sentence, MN // DE should MN // (DEF)
So: (MMfirstWOMENfirstN) // (DEF)
Derived from: MOTHERfirstWOMENfirst // (DEF)
c) Proven in sentence b.
Verse 6. Let two parallelograms ABCD and ABEF lie on two distinct planes. Let I, J, K in order be the centroids of triangles ADF, ADC, BCE. Prove that: (IJK) // (CDFE).
The answer
We have: CD // EF // AB so CD and EF are coplanar.
Let M, N be the midpoints of DF and CD, respectively.
Then, since I, J are the centroids of triangles ADF and ADC, respectively, so:
⇒ IJ // MN IJ // (CDEF)
Otherwise, let P be the midpoint CE.
Then:
We have: ABCD, ABEF are parallelograms, so CDEF is also a parallelogram.
Then for M, P is the midpoint of two opposite sides of parallelogram CDEF
Should: MP // CD //AB
Infer: IK // MP // AB
Hence: ABPK is also a parallelogram
We have:
Infer: IK // MN
Then: IK // (CDEF)
So: (IJK) // (CDEF)
Verse 7. Let S.ABCD pyramid with base is a parallelogram with center O. Let M, N, P be the midpoints SA, BC, CD respectively.
a) Find the intersection of two planes (SAD) and (MOP).
b) Let E be the midpoint of SC and I the point on the side SA satisfying AI = 3IS. Find K = IE ∩ (ABC) and H = AB ∩ (EIN). Calculate the ratio
.
c) Let G be the centroid of triangle SBC. Find the cross-section of the pyramidal S.ABC cut by (IMG).
The answer
a) We have: O, P are the midpoints of AC and CD, respectively, so OP // AD
Inferred: OP // (SAD)
Then: the intersection of (SAD) and (OMP) is a line through M and is parallel to AD and intersects SD at the midpoint Q of SD.
b) Consider the plane (SAC) with:
Inference: IE cuts AC at K
Then: K = IE (ABC)
Applying Menelaus’ theorem in triangle SAC with three collinear points K, E, I has:
Applying Menelaus’ theorem in triangle ABC with three collinear points K, N, H has:
Then:
c) We see, the plane (IMG) is also the plane (SAG).
Since: G is the centroid of triangle SBC and N is the midpoint of BC
So: (IMG) (SBC) = SN
So: The cross-section of the pyramid S.ABC cut by (IMG) is triangle SAN
Verse 8. Let the pyramid S.ABCD have base ABCD as a parallelogram with center O. Let M, N be the midpoints of SA and CD, respectively. Let E be the intersection of AD and (BMN), I be the midpoint of ME and G = AN ∩ BD.
a) Find the point E and the intersection F of SD with the plane (BMN). Prove: FS = 2FD.
b) Prove: FG // (SAB) and (CDI) // (SAB).
c) Let H be the intersection of MN and SG. Proof: OH // GF.
The answer
a) In the plane (ABCD) stretched, BN intersects the line AD at E.
Then: E is the intersection of (BMN) with AD
Let F be the intersection of ME with SD.
Then: F is the intersection of SD with (BMN)
Because:
So: D is mid point of segment AE
From this, it follows that: SD and EM are the medians of the triangle SAE .
Infer: F is the centroid of triangle SAE
So: FS = 2FD
b) Triangle DGN and triangle BGA are similar, so
Thence inferred:
Should: FG // SB FG // (SAB)
We have: CD // AB and DI // MA
From that, infer: (CDI) // (SAB)
c) We have: G is the centroid of triangle ACD.
Applying Menelaus’ theorem to the triangle SAG with three collinear points M, H, N we have:
We also have:
According to the above proof, we also have: GF // SB
So: OH // GF
Verse 9. Let the pyramid S.ABCD have base ABCD as a parallelogram with center O. Let M be the midpoint of SC, N be the point on diagonal BD such that BD = 3BN.
a) Determine the intersection of the plane (SCD) and (SAB) and find T = DM ∩ (SAB). Calculate
.
MS:
b) Call K = AN ∩ Prove that: MK // (SBD).
c) Let I = AN DC, L = IM Calculate the ratio
and .
MS:
The answer
a) The planes (SAB) and (SCD) contain two parallel lines, AB and CD, respectively, so their intersection is the line d through S and d // AB // CD.
In the plane (SCD) stretched DM intersects d at T.
Then: T d T ∈ (SAB)
So: T = DM ∩ (SAB)
Because CD // DT, two triangles MCD and MST are similar.
Therefore:
So:
b) Because:
Therefore: N is the centroid of triangle ABC
It follows that K is the midpoint of BC .
Which leads to MK being the midline of the triangle SBC.
Should: MK // SB MK // (SBD)
c) Triangle IKC and triangle IAD are similar, so
Applying Menelaus’ theorem to triangle SCD with three points I, M, L collinear we have:
Applying Menelaus’ theorem to triangle IDL with three collinear points S, M, C we have:
Thence inferred:
Let h, k be the lengths of the altitudes of the triangles IKM and IAL drawn from M and L, respectively.
It is easy to see that:
So:
Verse 10. Given two squares ABCD and ABEF in two distinct planes. On the diagonals AC and BF, take the points M and N, respectively, such that AM = BN. Lines parallel to AB drawn from M, N intersect AD and AF at M’ and N’, respectively. Prove that:
a) (ADF) // (BCE)
b) (CDF) // (MM’N’N)
The answer
a) We have:
b) We have:
(first)
We also have:
(2)
And from the assumption we have:
(3)
From (3) infer: M’N’ // DF.
We also have: MM’ // NN’ // DC // FE
So: (CDF) // (MM’N’N)
Verse 11. Let ABC.A’B’C’ prism. Let I, J, K be the centroids of triangles ABC, ACC’, A’B’C’, respectively. Prove that: (IJK) // (BCC’B’) and (A’JK) // (AIB’).
The answer
+) Let M, N, P be the midpoints of BC, CC’ and B’C’, respectively.
According to the properties of the centroid of the triangle, we have:
Quadrilateral AMP’A’ is a parallelogram and has:
So: (IJK) // (BCC’B’)
+) Note that: plane (AIB’) is plane (AMB’), plane (A’JK) is plane (A’CP).
Because: AM // A’P, MB’ // CP (because quadrilateral B’MCP is a parallelogram)
So we have: (A’JK) // (AIB’).
Verse 12. Let the pyramid S.ABCD have base ABCD as trapezoid with big bottom AD and AD = 2BC, M ∈ BC. Let (P) be the plane passing through M, (P) // CD, (P) // SC, (P) intersects AD, SA, SB at N, P, Q, respectively.
a) Prove that: NQ // (SCD) and NP // SD.
b) Let H, K be the midpoints of SD and AD respectively. Prove that: (CHK) // (SAB).
The answer
a) From M we draw a line parallel to CD that cuts AD at N and cuts AB at E.
From M draw a line parallel to SC that intersects SB at Q.
Extend the SA cut EQ at P.
According to the construction, we infer: (EPN) // (SCD) and NQ ⊂ (EPN)
So: NQ // (SCD)
Because (P) // (SCD) and these two planes intersect (SAD) along the intersections NP and SD.
So we infer: NP // SD
b) We have: HK is the median of triangle SAD
Should: HK // SA (1)
Since K is the midpoint of AD, AK = BC.
Therefore: quadrilateral ABCK is a parallelogram
Infer: CK // AB (2)
From (1) and (2) infer: (CKH // (SAB).
Verse 13. Let the pyramid S.ABC have G as the centroid of triangle ABC. On segment SA, take two points M and N such that SM = MN = NA.
a) Prove that: GM // (SBC).
b) Let D be a point symmetrical to A through G. Prove that: (MCD) // (NBG).
c) Let H = DM ∩ (SBC). Prove that: H is the centroid of triangle SBC.
The answer
a) Let E be the mid point of BC. Then we have:
So: GM // (SBC)
b) From the assumption we deduce G, N are the midpoints of AD and AM respectively.
Hence: NG // MD (1)
Quadrilateral BDCG has E as the midpoint of the two diagonals, so it is a parallelogram.
Derived: BG // CD (2)
From (1) and (2) infer: (MCD) // (NBG)
c) We have: AE is the median of triangle SBC (3)
Applying Menelaus’ theorem to triangle SAE with three collinear points M, H, D we have:
(4)
From (3) and (4) infer: H is the centroid of triangle SBC.
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