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As shown in the figure, P is a point on the diagonal AC of ABCD with side length 1 (P does not coincide with a and C), and point E is on the ray BC, and PE = Pb. (1) verification: PE = PD; (2) PE ⊥ PD
It is proved that: (1) let GF ∥ AB cross AD and BC to G and f respectively through point P. as shown in the figure, ∵ quadrilateral ABCD is a square, ∵ quadrilateral AbfG and quadrilateral gfcd are rectangles, ∵ AGP and ≌ PFC are isosceles right triangles, ∵ GD = FC = FP, GP = Ag = BF, ∠ PGD = ∠ PFE = 90 degrees, and ∵ Pb = PE, ≌ BF = Fe, ≌ GP = Fe, ≌ EFP ≌ PGD (SAS) ≌ PE = PD; (2) ≌ EFP ≌ In the second proof, it is proved that: (1) ABCD is a square, AC is a diagonal, BC is DC, BCP is DCP 45, PC is PC, PBC is PDC & nbsp; (SAS) ∵ Pb = PD, ∠ PBC = ∠ PDC. Also ∵ Pb = PE, ∵ PE = PD; (2) ∵ Pb = PE, ∵ PBE = ∠ PEB, ∵ PEB = ∠ PDC, ∵ PEB + ∠ PEC = ∠ PDC + ∠ PEC = 180 °, ∵ DPE = 360 ° - (∠ BCD + ∠ PDC + ∠ PEC) = 90 ° and ⊥ PE ⊥ PD
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- 1. As shown in Figure 1, it is known that P is a point on the diagonal AC of square ABCD (not coincident with a and C), PE ⊥ BC at point E, PF ⊥ CD at point F. (1) verification: BP = DP; (2) as shown in Figure 2, if quadrilateral PECF rotates counterclockwise around point C, is there always BP = DP in the process of rotation? If so, please prove it; if not, please explain it with counter examples; (3) try to select two vertices of square ABCD and connect them with the two vertices of quadrilateral PECF respectively, so that the length of the two segments is always equal when quadrilateral PECF rotates counterclockwise around point C, and prove your conclusion
- 2. As shown in Figure 1, it is known that P is a point on the diagonal AC of square ABCD (not coincident with a and C), PE ⊥ BC at point E, PF ⊥ CD at point F. (1) verification: BP = DP; (2) as shown in Figure 2, if quadrilateral PECF rotates counterclockwise around point C, is there always BP = DP in the process of rotation? If so, please prove it; if not, please explain it with counter examples; (3) try to select two vertices of square ABCD and connect them with the two vertices of quadrilateral PECF respectively, so that the length of the two segments is always equal when quadrilateral PECF rotates counterclockwise around point C, and prove your conclusion
- 3. It is known that P is a point AC on the diagonal of square ABCD (not coincident with a and C), PE is perpendicular to point E, PF is perpendicular to point F 1, and CD is perpendicular to point F 1 It is known that P is a point AC (not coincident with a and C) on the diagonal of square ABCD, PE is perpendicular to point E, PF is perpendicular to point F 1. Verify BP = DP 2. If the quadrilateral PECF rotates counterclockwise around point C, is there BP = DP in the process of rotation? If so, please prove it, if not, please use counter examples Try to select two vertices of square ABCD and connect them with the two vertices of quadrilateral PECF, so that the length of the two segments is equal when the quadrilateral PECF rotates counterclockwise around point C, and prove your conclusion Thank you on your knees
- 4. The side length of square ABCD is a, PA ⊥ plane ABCD, PA = a, m, n are the midpoint of PD, Pb respectively, then the cosine value of the angle formed by the line am and cn is
- 5. As shown in the figure, a point P, PE ⊥ ad in the square ABCD is equal to E. if Pb = PC = PE = 5, then the side length of the square is______ ．
- 6. It is known that: as shown in the figure, in square ABCD, point E is a point on the side of AD, and the intersection diagonal of CE is BD at point P, PE = AE. (1) verification: CE = 2ed. (2) when Pb = 6cm Please teach me the way I understand, Just a second question
- 7. Let p be any point on the inferior arc ad of circumscribed circle of square ABCD, then the ratio of PA + PC to Pb is______ ．
- 8. The center of square ABCD is O, the area is 1989cm2. P is a point in the square, and ∠ OPB = 45 °, PA: Pb = 5:14=______ ．
- 9. There is a point P in the square ABCD with side length 2. Find the minimum value of PA + Pb + PC. please write the procedure. 7. Ab There is a point P in the square ABCD with side length 2. Find the minimum value of PA + Pb + PC. please write the procedure 7. AB and AC are the diameter chord of circle O respectively, D is the point on inferior arc AC, De is perpendicular to ab at point h, intersecting circle O at point E, intersecting circle AC at point F, P is a point on extended line of ED
- 10. As shown in the figure, in isosceles trapezoid ABCD, ad ‖ BC, ∠ B = 60 °, ad = ab. points E and F are on AD and ab respectively, AE = BF, DF and CE intersect at P, then ∠ DPE=______ Degree
- 11. As shown in the figure, P is a moving point on the diagonal AC of the square ABCD with side length 1 (P does not coincide with A. C), and point E is on the ray BC And PE = PB Verification (1) PE vertical PD Let AP = x △ PBE area be y Find out the function relation between Y and X, and write out the value range of X
- 12. In square ABCD, point P is a point on edge CD, and PE ⊥ dB, PF ⊥ Ca, perpendicular feet are points E and f respectively, What is the quantitative relationship between PE + PF and the diagonal length of a square
- 13. It is known that, as shown in the figure, P is a point inside the square ABCD, and there is a point e outside the square ABCD, satisfying that the angle Abe = angle CBP, be = BP, If PA: Pb = 1:2, angle APB = 135 degrees, what is the chord value of angle PAE?
- 14. As shown in the figure, P is a point in the square ABCD, and there is a point e outside the square ABCD, which satisfies ∠ Abe = ∠ CBP, be = BP,
- 15. Let p be a point in a square ABCD, △ ABP is congruent with △ CBE. If BP = a, let PE be the area of the square with side length
- 16. For a point in the square ABCD, rotate △ ABP clockwise 90 ° around B to the position of △ CBE, if BP = A. calculate the area of the square with PE as the side length Write every step in detail Using Pythagorean theorem
- 17. As shown in the figure, point P is a point in square ABCD. Rotate △ ABP around point B and coincide with △ BCP '. If BP = 3, find the length of PP ′ De
- 18. Point P is a point in the square ABCD, connecting PA, Pb and PC. rotate the triangle PAB 90 degrees clockwise around point B to the position of triangle p'cb Let AB be a and Pb be B (b)
- 19. P is a point in the square ABCD, and AP = 2. Rotate △ APB clockwise 60 ° around a to get △ AP ′ B ′. (1) make the rotated figure; (2) try to find the perimeter and area of △ app ′
- 20. In parallelogram ABCD, P is a point outside ad, and AP is vertical to PC, BP is vertical to PD, so find parallelogram ABCD as rectangle