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It is known that: as shown in the figure, P is the interior point of the square ABCD, ∠ pad = ∠ PDA = 15 °. Verification: △ PBC is an equilateral triangle
It is proved that: square ABCD, AB = CD, bad = ∠ CDA = 90 °, ? pad = PDA = 15 °, PA = PD, ∠ PAB = ∠ PDC = 75 °, make △ DGC and △ ADP congruent in the square, ? DP = DG, ∠ ADP = ∠ GDC = ∠ DAP = ∠ DCG = 15 °, ? PDG = 90 ° - 15 ° = 60 °, Δ PDG
PD ⊥ face ABCD, ad ⊥ DC, ad ∥ BC, PD: DC: BC = 1:1: √ 2. (1): calculate the angle between Pb and plane PDC. (2): find the tangent of dihedral angle d-pbc Twenty is yours
(1) 45 degrees
P is any point in the rectangle ABCD. Connect PA, Pb, PC and PD to get △ PAB, △ PBC, △ PCD and △ PDA. If their areas are S1, S2, S3 and S4, then S1 * S3 = S2 * S4. Is this right
Error. Should be S1 + S3 = S2 + S4
Given the square ABCD, please find out the point P that meets the following conditions: the connecting line PA, Pb, PC, and PD, △ PAB, △ PBC, △ PCD and △ PDA are isosceles triangles. There are eight such points. Draw a picture A D B C Say it. If you draw eight bonus points
How do I think there are countless? Find out the center of gravity m of square ABCD, make a straight line perpendicular to the plane ABCD through M, then any point P on this line satisfies the condition
Taking the side length BC of the square ABCD as the edge, an equilateral triangle PBC is made in the square to connect PA and PD
Two
Given that the BC side of the square ABCD is the length of the side, make an equilateral △ PBC in the square to connect PA and PD 1. Find the degree of ∠ ADB and ∠ APD 2. If AB = 2, find s △ ABP
1 ∠ADB＝45º.∠APD＝360º-60º-2×[180º-30º]/2＝150º
2 s △ ABP = (√ 3 / 4) AB 2 = √ 3 (area unit)
The area of the square ABCD is 1, the point P is in the square, and △ PBC is an equilateral triangle
The side length of 1 is 1,
∵ △ BPC area = √ 3 / 4
Area of △ cpd = 1 / 4,
Area of △ BCD = 1 / 2,
The area of △ PBD = △ BPC + cpd - △ BCD = √ 3 / 4 + 1 / 4-1 / 2 = = √ 3 / 4-1 / 4
It is known that: as shown in the figure, P is the interior point of the square ABCD, ∠ pad = ∠ PDA = 15 °. Verification: △ PBC is an equilateral triangle
It is proved that: square ABCD, AB = CD, bad = ∠ CDA = 90 °, ? pad = PDA = 15 °, PA = PD, ∠ PAB = ∠ PDC = 75 °, make △ DGC and △ ADP congruent in the square, ? DP = DG, ∠ ADP = ∠ GDC = ∠ DAP = ∠ DCG = 15 °, ? PDG = 90 ° - 15 ° = 60 °, Δ PDG
A point P in the square ABCD and the angle pad is equal to the angle PDA is equal to 15 degrees
Using the same method
The angle p'bc = angle p'cb = 60 degrees, p 'is the intersection point of BP' and CP '
Delta p'bc equilateral triangle
Angle p'ba = 30 degrees
AB=BC=BP'
Δ p'ba is an isosceles triangle
Angle p'ab = 75 degrees
Angle p'ad = 15 degrees
The same angle p'da = 15 degrees
There is only one intersection point of the 15 degree edges at a and D
So p 'coincides with P
Therefore, the original proposition holds
There is a point P in the square, PA = Pb, and angle PAB = angle PBA = 15 degrees. It is proved that the triangle PCD is a regular triangle
Master style is extraordinary!
It is wise to use the same method to prove this question
It is proved that CD is an equilateral triangle MCD in a square, connecting Ma and MB
Then the angle MCD = angle MDC = 60 degrees, angle ADM = angle BCM = 30 degrees
MC = MD = CD = ad = BC, triangle BCM and triangle ADM are isosceles triangles
So the angle mad = angle MBC = (180 degrees - 30 degrees) / 2 = 75 degrees,
Then the angle mAb = angle MBA = 15 degrees, and angle PAB = angle PBA = 15 degrees
So BP and BM are on the same line, AP and am are on the same line,
According to "two straight lines intersect, only one intersection", points P and M coincide,
That is, the triangle PCD is a regular triangle
RELATED INFORMATIONS
- 1. The bottom surface of four pyramids P-ABCD is square, and the bottom surface of PB ⊥ ABCD proves that no matter how the height of four pyramids PB changes, the surface PAD and the surface PCD cannot be vertical
- 2. As shown in the figure, ∠ AOB = 30 ° OC bisection ∠ AOB, P is a point on OC, PD ‖ OA intersects ob at point D, PE ⊥ OA at e, OD = 4cm, then PE ⊥ OA is at point E=______ .
- 3. Known: as shown in the figure, find a point P on the straight line Mn, so that the distance between the point P and the two sides of ∠ AOB is equal (the method is required to be written, the drawing trace shall be retained, and the conclusion shall be written)
- 4. In the plane rectangular coordinate system, the image of function y = - 3 / 4x + 6 intersects X axis and Y axis at point a and B respectively, line BC and X axis intersect at C, and point C is the midpoint of line OA Find out whether there is a point m on the straight line AB so that △ BCM is an isosceles triangle. If there is, find out the coordinates of point M
- 5. As shown in the figure, the ray OA, ob, OC, OD have common endpoint o, and ∠ AOB = 90 °, COD = 90 °, AOD = 5 4 ﹤ AOC. calculate the degree of ﹤ BOC
- 6. As shown in the figure, OA, OB and OC are the radii of circle O. if ∠ AOB = 2 ∠ BOC, please judge whether ∠ ACB = 2 ∠ BAC is tenable. Why?
- 7. The right angle vertex P of the triangle ruler slides on the ray OM, and the two right angle sides intersect OA and ob at point CD respectively Conjecture the quantitative relationship between PC and PD
- 8. As shown in the figure, the side of the cone is cut along OA, and the fan-shaped OAB is expanded into a plane figure (1) What is the relationship between the length of the sector arc AB and the length of the circle at the bottom of the cone? What is the position relationship between point a and point B on the side of the cone? (2) If the angle ∠ AOB = 90 °, what is the relationship between the radius r of the cone bottom circle and the radius r of fan-shaped OAB (i.e. OA or OB)? (3) If point a moves on the side of the cone and then returns to its original position, how should the shortest distance of point a be designed? If R 2 = 0.5, ∠ AOB = 90 °, find the shortest distance of point a
- 9. As shown in the figure, it is known that ∠ AOB = 30 ° and the point P is inside ∠ AOB, Op = 6. If there is a moving point m on OA and a moving point n on ob, then the minimum perimeter of △ PMN is______ .
- 10. In the triangle ABC, if AC = radical 2, BC = radical 7, ab = 3, then cosa =?
- 11. Ray OA, ob, OC, OD have a common endpoint o, and OA is perpendicular to ob, OC is perpendicular to OD, ∠ AOD = 5 / 4 ∠ AOC, calculate the degree of ∠ BOC ∠ AOD = 5 / 4 ∠ AOC! No, ∠ AOD = 5 / 4 ∠ BOC
- 12. As shown in Fig
- 13. It is proved that the coordinates of the point P which makes the vector AP * vector BP get the minimum value is [emergency] given that the point a (2,0), B (4,0), and the moving point P move in the parabola y ^ 2 = - 4x, it is proved that the coordinate of the point P which makes the vector AP * vector BP get the minimum value is the coordinate origin. Please give the detailed process,
- 14. Mathematical problem: introduce two tangent lines PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1 1, introduce two tangent PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1, and the tangent points are a, B, and ∠ APB = 60 degrees, then the trajectory equation of the moving point P is________ 2. Let P (x, y) be a moving point on the circle x ^ 2 + (Y-1) ^ 2 = 1. If the inequality x + y + C > 0 holds, then the value range of C_____________ 3. We know that the equation of the line where the edge ab of the equilateral △ ABC is located is x √ 3 + y = 0, and the coordinates of point C are (1, √ 3) 4. It is known that the circle M: x ^ 2 + (Y-2) ^ 2 = 1, q is the moving point on the x-axis, QA and QB tangent the circle m to two points a and B respectively (1) (liberal arts) if | ab | = 4 √ 2 / 3, find the equation of line MQ (Science) prove that the straight line AB always crosses a fixed point (2) Find the locus equation of midpoint P of moving string ab It's better to analyze it
- 15. As shown in the figure, lead the tangent PA, Pb of the circle through a point p outside ⊙ o with radius of 6cm, connect Po, intersect ⊙ o at F, make tangent of ⊙ o through F, intersect PA and Pb at D and e respectively, if Po = 10cm, ﹤ APB = 40 ° Find: (1) the circumference of △ ped; (2) the degree of ∠ doe
- 16. Pass through the plane α of the triangle ABC and a point p outside the plane α of △ ABC, make Po ⊥ α, and connect PA, Pb, PC with the perpendicular foot of O 2. If PA = Pb = PC, then o is △ ABC__ heart
- 17. As shown in the graph, P is a point in the equilateral △ ABC, PA = 6, Pb = 8, PC = 10, if the point P 'is a point outside △ ABC, but △ p'ab is congruent with △ PAC The distance between point P and point P 'and the degree of ∠ APB
- 18. As shown in the figure, BC is the diameter of ⊙ o, P is a point on ⊙ o, a is the midpoint of arc BC, ad ⊥ BC, perpendicular foot is D, Pb intersects AD and AC respectively at points e, F, AE and be? Why? It stops after 20 minutes
- 19. It is known that P is a point in the plane of triangle ABC. If vector CB = γ vector PA + vector Pb, γ belongs to R, then point P must be in the plane A. Inside the triangle ABC B. On the line where the AC side is located C. On the line where AB side is located D. On the line where the BC side is located Why?
- 20. Plane PAC ⊥ plane ABC, PA = Pb = PC, verify ab ⊥ BC