这道题怎么解决?---**Textual Information:** 1. (2020・孝感)如图,点 E 在正方形 ABCD 的边 CD 上,将△ADE 绕点 A 顺时针旋转 90° 到△ABF 的位置,连接 EF,过点 A 作 EF 的垂线,垂足为点 H,与 BC 交于点 G.若 BG=3, CG=2, 则 CE 的长为 ( ) Options: A. $\frac{5}{4}$ B. $\frac{15}{4}$ C. 4 D. $\frac{9}{2}$ **Chart/Diagram Description:** * **Type:** Geometric figure illustrating a square and related points and lines. * **Main Elements:** * **Shape:** A square labeled ABCD. * **Points:** Vertices of the square are labeled A (top-left), B (bottom-left), C (bottom-right), and D (top-right). Point E is on the side CD. Point F is located such that triangle ABF is the result of rotating triangle ADE clockwise by 90 degrees around point A. Point G is on the side BC. Point H is on the line segment EF and also on the line segment AG. * **Lines:** The sides of the square AB, BC, CD, DA are drawn. Line segments AE, DE, AF, BF, EF, AG are drawn. The line segment AG is drawn passing through H and intersecting BC at G. There is a right angle symbol at H where AG intersects EF, indicating that AG is perpendicular to EF. * **Labels:** All points A, B, C, D, E, F, G, H are labeled. * **Relative Position and Direction:** The square ABCD is oriented with AD and BC vertical, AB and CD horizontal. E is on CD. F is located such that ABF is a rotation of ADE (meaning AB aligns with AD after rotation, and F aligns with E). G is on BC between B and C. H is the intersection of AG and EF. AG is shown passing through the interior of the square and triangle ABF. Based on the description and the diagram: - Since E is on CD and the square is ABCD, AD is a side length. DE is a part of CD. - Triangle ADE is rotated clockwise by 90 degrees around A to get triangle ABF. This means: - AD rotates to AB. - AE rotates to AF. - DE rotates to BF. - Angle DAE equals angle BAF. - Angle D equals angle ABF (both 90 degrees). - Lengths AD=AB (side of square), DE=BF, AE=AF. - Point G is on BC. - AG is perpendicular to EF at H. - BG=3 and CG=2 are given. Since G is on BC, BC = BG + CG = 3 + 2 = 5. Thus, the side length of the square is 5. AD = AB = BC = CD = 5. - E is on CD, so DE + CE = CD = 5. We need to find CE. If we find DE, we can find CE. - Since DE=BF and CD=BC=5, and G is on BC, BF could be related to BG or CG or the side length. However, F is not necessarily on BC. F is obtained by rotating E. Since E is on CD, F will be such that BF = DE and AB=AD. As E is on CD, D is at (0,5) and A is at (0,0) if B is at (5,0) and C at (5,5) (assuming A is origin, AB along x-axis, AD along y-axis). Then AB is along the x-axis. Rotating E(xE, 5) 90 degrees clockwise around A(0,0) means the new point F(xF, yF) will have coordinates (5, -xE). So F is at (5, -DE). B is at (5,0). This means F is vertically below B, and BF = DE. Yes, F is on the line x=5, which is the line BC. The coordinate system assumption might be wrong based on the diagram. Let's use geometric properties. Rotate D by 90 degrees clockwise around A gives B. Rotate E by 90 degrees clockwise around A gives F. So triangle ADE is congruent to triangle ABF. DE = BF and AE = AF. Given BG=3 and CG=2, the side length of the square BC = BG + CG = 3 + 2 = 5. So AB = BC = CD = DA = 5. Since DE = BF and E is on CD, DE is between 0 and 5. So BF is between 0 and 5. In the diagram, F is shown such that B is between C and F or F is on the extension of CB. However, the rotation maps E on CD to F. D maps to B. So the image of the line segment CD after rotation is a line segment passing through B and perpendicular to AB. Since CD is parallel to AB and perpendicular to AD, rotating CD 90 degrees clockwise around A means the image of CD is parallel to AD and perpendicular to AB. Since D rotates to B, the image line passes through B and is parallel to AD. This is the line containing BC. As E is on CD, F is on the line containing BC. Also, AE rotates to AF, so triangle AEF is an isosceles triangle with AE=AF. Since F is the image of E after rotating 90 degrees around A, angle EAF = 90 degrees and AE=AF. AG is perpendicular to EF. In triangle AEF, which is an isosceles right triangle if AE=AF, AG being perpendicular to EF is expected. AG is an altitude in triangle AEF from A to EF. Consider the line AG intersecting BC at G. We are given BG=3, CG=2. BC=5. Let CE = x. Then DE = CD - CE = 5 - x. Since DE = BF (from rotation), BF = 5 - x. F is on the line containing BC. B is on this line. BF = 5-x. There are two possibilities for the position of F on the line containing BC relative to B: F could be such that B is between C and F, or B is between C and B (which is G) and F is further away. Or F is on the line containing BC. Since triangle ADE is rotated to triangle ABF, point F is the image of E. E is on CD. Let's set up a coordinate system with A at (0,0), B at (5,0), C at (5,5), D at (0,5). E is on CD, so E has coordinates (xE, 5) where 0 <= xE <= 5. Let DE = d. Then E is at (5-d, 5). Rotating E(5-d, 5) by 90 degrees clockwise around (0,0) gives F. Rotation matrix for 90 degrees clockwise is [[0, 1], [-1, 0]]. So F = (0*(5-d) + 1*5, -1*(5-d) + 0*5) = (5, -(5-d)) = (5, d-5). So F is at (5, d-5). Point B is at (5,0). Point C is at (5,5). The line containing BC is the line x=5. F is on this line. BF is the distance between (5,0) and (5, d-5), which is |0 - (d-5)| = |5-d|. Since DE = d, and BF=DE, BF = d. So d = |5-d|. Since E is on CD, 0 <= d <= 5. So 5-d >= 0. Therefore, d = 5-d, which means 2d=5, d=2.5. So DE=2.5 and CE = 5-2.5 = 2.5. In this case, BF = 2.5. Since B is at (5,0) and F at (5, d-5), i.e., (5, -2.5), F is at (5, -2.5). The diagram shows F below B. Let's re-check the rotation. Rotating D(0,5) 90 degrees clockwise around A(0,0) gives B(5,0). This is correct. Rotating E(5-d, 5) 90 degrees clockwise around A(0,0) gives F(5, d-5). Yes. F is at (5, d-5). B is at (5,0). The y-coordinate of F is d-5. Since 0<=d<=5, d-5 is between -5 and 0. So F is below or at B. The distance BF is |0 - (d-5)| = |5-d|. Since d is DE, BF = DE = d. So d = |5-d|. Since 0<=d<=5, 5-d>=0. So d = 5-d, which gives d=2.5. So DE=2.5 and CE=2.5. BF=2.5. F is at (5, -2.5). However, the problem states BG=3 and CG=2, so BC=5, which is consistent with the side length being 5. G is on BC, so G is at (5, yG) where 0 <= yG <= 5. Since B is at (5,0) and C at (5,5), BG = yG, CG = 5-yG. Given BG=3, CG=2, then yG = 3. So G is at (5,3). The line AG passes through A(0,0) and G(5,3). The equation of line AG is y = (3/5)x. The line segment EF connects E(5-d, 5) and F(5, d-5). The slope of EF is ( (d-5) - 5 ) / ( 5 - (5-d) ) = (d-10) / d. The line AG is perpendicular to EF. So the product of their slopes is -1. Slope of AG is 3/5. Slope of EF is (d-10)/d. (3/5) * ((d-10)/d) = -1 3(d-10) = -5d 3d - 30 = -5d 8d = 30 d = 30/8 = 15/4. d is DE. So DE = 15/4. CE = CD - DE = 5 - 15/4 = 20/4 - 15/4 = 5/4. Let's check the options. A. 5/4 B. 15/4 C. 4 D. 9/2 Our calculated value for CE is 5/4, which matches option A. Let's quickly verify if the calculated value makes sense with the diagram. DE = 15/4 = 3.75. Since CD=5, E is on CD. 0 <= DE <= 5. 3.75 is within this range. CE = 5/4 = 1.25. BF = DE = 15/4 = 3.75. F is at (5, d-5) = (5, 15/4 - 5) = (5, 15/4 - 20/4) = (5, -5/4). So F is at (5, -1.25). B is at (5,0). BF is the distance between (5,0) and (5, -5/4), which is |0 - (-5/4)| = 5/4. Wait, there is a contradiction. BF=DE=d, and BF = |5-d|. So d = |5-d|. This led to d=2.5. But the perpendicularity condition led to d=15/4. Let's re-evaluate the calculation of F's coordinates. A=(0,0), D=(0,5), E is on CD. Let E=(xE, 5). DE is the distance between (0,5) and (xE, 5), which is xE. So d = xE. E is at (d, 5). No, this is wrong. If A is (0,0), B(5,0), C(5,5), D(0,5). E is on CD. CD is the line segment from (0,5) to (5,5). E has coordinates (xE, 5), where 0 <= xE <= 5. DE is the distance between (xE, 5) and (0,5), which is xE. Let DE = d. Then E is at (d, 5). CD length is 5. CE = 5-d. Rotate E(d, 5) 90 degrees clockwise around A(0,0). Rotation matrix is [[0, 1], [-1, 0]]. F = (0*d + 1*5, -1*d + 0*5) = (5, -d). So F is at (5, -d). B is at (5,0). C is at (5,5). BF is the distance between (5,0) and (5, -d), which is |0 - (-d)| = |d|. Since d=DE, and E is on CD, 0<=d<=5. So BF = d. This is consistent with BF=DE. A is at (0,0). G is on BC. BC is the line x=5, 0<=y<=5. G is on BC. BG=3, CG=2. Since B is at (5,0) and C at (5,5), G is at (5,3). Line AG passes through A(0,0) and G(5,3). Slope of AG is (3-0)/(5-0) = 3/5. Equation of line AG is y = (3/5)x. Line segment EF connects E(d, 5) and F(5, -d). Slope of EF is (-d - 5) / (5 - d). AG is perpendicular to EF. So the product of slopes is -1. (3/5) * ((-d - 5) / (5 - d)) = -1 3(-d-5) = -5(5-d) -3d - 15 = -25 + 5d -15 + 25 = 5d + 3d 10 = 8d d = 10/8 = 5/4. d is DE. So DE = 5/4. CE = CD - DE = 5 - 5/4 = 20/4 - 5/4 = 15/4. Let's check the options again. A. 5/4 (This was our initial calculation of CE when using the incorrect formula for F) B. 15/4 (This is our new calculation for CE) C. 4 D. 9/2 = 4.5 Our new calculation for CE is 15/4, which matches option B. Let's double check the coordinates and calculations. Square ABCD, side length 5. A=(0,0), B=(5,0), C=(5,5), D=(0,5). E is on CD. CD is the segment from (0,5) to (5,5). Let DE = d. E is at (d, 5). (0<=d<=5). Rotate E(d,5) 90 deg clockwise around A(0,0) to F. Rotation matrix [[cos(-90), -sin(-90)], [sin(-90), cos(-90)]] = [[0, 1], [-1, 0]]. F = (0*d + 1*5, -1*d + 0*5) = (5, -d). F is at (5, -d). Since 0<=d<=5, -d is between -5 and 0. This means F is on the line x=5, below B (at (5,0)) or at y=-5 (if d=5, E=C, F=(5,-5)). G is on BC. B=(5,0), C=(5,5). BG=3, CG=2. BG+CG=5=BC. G is at (5, 3). Line AG passes through A(0,0) and G(5,3). Slope mAG = (3-0)/(5-0) = 3/5. Line EF connects E(d,5) and F(5,-d). Slope mEF = (-d-5)/(5-d). AG is perpendicular to EF, so mAG * mEF = -1. (3/5) * ((-d-5)/(5-d)) = -1 3(-d-5) = -5(5-d) -3d - 15 = -25 + 5d -15 + 25 = 5d + 3d 10 = 8d d = 10/8 = 5/4. d is DE. DE = 5/4. CE = CD - DE = 5 - 5/4 = 20/4 - 5/4 = 15/4. The steps seem correct now. Let's re-check the initial coordinate system where D=(0,5), C=(5,5), A=(0,0), B=(5,0). This assumes AD is along y-axis and AB along x-axis. E is on CD. CD is from (0,5) to (5,5). So E is at (xE, 5), where 0<=xE<=5. DE is the distance from (xE, 5) to (0,5), which is xE. Let DE=d. Then E is at (d, 5). Rotate E(d, 5) 90 degrees clockwise around A(0,0). Rotation matrix is [[0, 1], [-1, 0]]. F = (0*d + 1*5, -1*d + 0*5) = (5, -d). So F is at (5, -d). B is at (5,0). This means BF is vertical distance |0 - (-d)| = |d|. Since d=DE>=0, BF=d. This matches the property of rotation: DE=BF. G is on BC. B=(5,0), C=(5,5). G is at (5, yG). BG=|yG-0|=yG. CG=|5-yG|. BG=3, CG=2. So yG=3. G is at (5,3). Line AG passes through A(0,0) and G(5,3). Slope mAG = 3/5. Line EF connects E(d,5) and F(5,-d). Slope mEF = (-d-5)/(5-d). AG is perpendicular to EF, so mAG * mEF = -1. (3/5) * ((-d-5)/(5-d)) = -1 3(-d-5) = -5(5-d) -3d - 15 = -25 + 5d 10 = 8d d = 10/8 = 5/4. So DE = 5/4. CE = CD - DE = 5 - 5/4 = 15/4. The coordinate setup and calculation seem consistent now and lead to CE=15/4. Let's confirm the options again. A. 5/4 B. 15/4 C. 4 D. 9/2 Our result is 15/4, which is option B. Double check the problem statement and diagram. Square ABCD. E on CD. Rotate triangle ADE around A 90 degrees clockwise to triangle ABF. Connect EF. Line through A perpendicular to EF at H, intersects BC at G. BG=3, CG=2. Find CE. Diagram shows A top-left, B bottom-left, C bottom-right, D top-right. This orientation is different from the coordinate system assumed above. Let's re-orient the coordinate system based on the diagram. A is top-left. Let A=(0,0). AB is horizontal. AD is vertical. A=(0,0), B=(s,0), C=(s,s), D=(0,s). Where s is the side length of the square. G is on BC. BC is the line segment from (s,0) to (s,s). G is on BC. Let G=(s, yG). BG is the distance between (s,0) and (s,yG), which is |yG-0|=yG. CG is the distance between (s,s) and (s,yG), which is |s-yG|. Given BG=3 and CG=2. So yG=3 and s-yG=2. s-3=2, so s=5. Side length is 5. A=(0,0), B=(5,0), C=(5,5), D=(0,5). This is the same coordinate system as before. E is on CD. CD is the line segment from (0,5) to (5,5). E=(xE, 5), where 0<=xE<=5. DE is the distance from (0,5) to (xE, 5), which is xE. Let DE=d. E is at (d, 5). Rotate E(d, 5) 90 degrees clockwise around A(0,0) to F. Rotation matrix [[0, 1], [-1, 0]]. F = (0*d + 1*5, -1*d + 0*5) = (5, -d). G is at (5,3). Line AG passes through A(0,0) and G(5,3). Slope mAG = 3/5. Line EF connects E(d,5) and F(5,-d). Slope mEF = (-d-5)/(5-d). AG is perpendicular to EF. mAG * mEF = -1. (3/5) * ((-d-5)/(5-d)) = -1 3(-d-5) = -5(5-d) -3d - 15 = -25 + 5d 10 = 8d d = 10/8 = 5/4. d is DE. DE = 5/4. CE = CD - DE = 5 - 5/4 = 15/4. The calculations consistently give CE = 15/4. Let's think about the positions of points in the diagram. A is top-left, B bottom-left, C bottom-right, D top-right. AB is bottom edge, BC right edge, CD top edge, DA left edge. A=(0,5), B=(5,5), C=(5,0), D=(0,0). Side length is 5. E is on CD. CD is from (0,0) to (5,0). E is at (xE, 0), where 0<=xE<=5. DE = xE. Let DE=d. E is at (d, 0). Rotate E(d, 0) 90 degrees clockwise around A(0,5). Center of rotation is A(0,5). Point to rotate is E(d,0). Vector AE = (d-0, 0-5) = (d, -5). Rotate this vector 90 degrees clockwise: (y, -x) = (-5, -d). Translate back by A's coordinates: F = (0+(-5), 5+(-d)) = (-5, 5-d). So F is at (-5, 5-d). B is at (5,5). This location of F seems inconsistent with the diagram. The diagram shows F near B, and possibly below or to the left of B. Let's try the first coordinate system orientation again, as it led to a valid answer option. A=(0,0), B=(5,0), C=(5,5), D=(0,5). This is A bottom-left, B bottom-right, C top-right, D top-left. This also doesn't match the diagram. Let's try A=(0,0), B=(5,0), C=(5,-5), D=(0,-5). This makes ABCD a square in the 4th quadrant. But AB is on the x-axis. Let's assume the diagram orientation is standard geometry: A=(0,5), B=(0,0), C=(5,0), D=(5,5). A=(0,5), B=(0,0), C=(5,0), D=(5,5). Side length 5. E is on CD. CD is from (5,5) to (5,0). E is at (5, yE), where 0<=yE<=5. DE is the distance between (5,yE) and (5,5), which is |yE-5| = 5-yE (since yE<=5). Let DE=d. Then yE=5-d. E is at (5, 5-d). Rotate E(5, 5-d) 90 degrees clockwise around A(0,5). Vector AE = (5-0, (5-d)-5) = (5, -d). Rotate vector AE 90 degrees clockwise: (y, -x) = (-d, -5). Translate back by A's coordinates: F = (0+(-d), 5+(-5)) = (-d, 0). So F is at (-d, 0). B is at (0,0). F is on the x-axis, to the left of B (or at B if d=0). G is on BC. BC is the line segment from (0,0) to (5,0). So BC is on the x-axis? No, BC is the segment from (0,0) to (5,0) in this coordinate system, which is the x-axis from x=0 to x=5. G is on BC. BG=3, CG=2. Length of BC is 5. B=(0,0), C=(5,0). G is on the segment [0,5] on the x-axis. Distance from B(0,0) to G(xG,0) is |xG|. Distance from C(5,0) to G(xG,0) is |xG-5|. If G is between B and C, then xG = BG = 3. CG = 5-3=2. So G is at (3,0). Line AG passes through A(0,5) and G(3,0). Slope mAG = (0-5)/(3-0) = -5/3. Line EF connects E(5, 5-d) and F(-d, 0). Slope mEF = (0 - (5-d)) / (-d - 5) = (d-5) / (-d-5) = (5-d) / (d+5). AG is perpendicular to EF. mAG * mEF = -1. (-5/3) * ((5-d)/(d+5)) = -1 -5(5-d) = -3(d+5) -25 + 5d = -3d - 15 5d + 3d = -15 + 25 8d = 10 d = 10/8 = 5/4. d is DE. DE = 5/4. CE = CD - DE. CD is the length of side CD, which is 5. E is on CD. CE = 5 - DE = 5 - 5/4 = 15/4. This third coordinate system orientation (A top-left, B bottom-left, C bottom-right, D top-right) gives the same result CE=15/4. This orientation matches the diagram visually. Let's assume this is the intended setup. Let's briefly try a geometric approach without coordinates to build confidence. Rotate triangle ADE about A 90 degrees clockwise to get triangle ABF. So AD rotates to AB, DE rotates to BF, AE rotates to AF. Angle D = Angle ABF = 90 degrees. DE = BF, AE = AF. Angle EAF = 90 degrees. AG is perpendicular to EF. In triangle AEF, AE=AF and angle EAF = 90 degrees. This is an isosceles right triangle. AG is the altitude from A to EF. In an isosceles right triangle, the altitude from the vertex with the right angle bisects the hypotenuse. So H is the midpoint of EF. Let M be the midpoint of EF. Then M=H, and AG passes through M and is perpendicular to EF. Let CE = x. Then DE = 5-x. BF = DE = 5-x. Let's use vectors or slopes in a coordinate system. We already did that. Consider similar triangles formed by the perpendicularity. AG is perpendicular to EF. Let's consider slopes again with A=(0,5), B=(0,0), C=(5,0), D=(5,5). E on CD. CD is x=5, 0<=y<=5. E=(5, yE). DE = 5-yE. CE=yE. Let CE=x. E=(5,x). DE=5-x. Rotate E(5,x) 90 deg clockwise around A(0,5). Vector AE = (5-0, x-5) = (5, x-5). Rotate vector AE 90 deg clockwise: (y, -x) = (x-5, -5). Translate back by A's coordinates: F = (0+(x-5), 5+(-5)) = (x-5, 0). F is at (x-5, 0). B is at (0,0). C is at (5,0). F is on the x-axis. If x=CE, then 0<=x<=5. Then x-5 is between -5 and 0. So F is on the x-axis between (-5,0) and (0,0), or at (-5,0) or (0,0). So F is on the line containing BC, to the left of B (or at B if x=5, E=C, F=(0,0)=B). Distance BF = |(x-5)-0| = |x-5|. Since x<=5, x-5<=0. So BF = -(x-5) = 5-x. From rotation, DE=BF. DE=5-x. So BF=5-x. This is consistent. G is on BC. B=(0,0), C=(5,0). G is at (xG, 0). BG=|xG|. CG=|xG-5|. BG=3, CG=2. Since BG+CG=5=BC, G is between B and C. So xG=3. G=(3,0). Line AG passes through A(0,5) and G(3,0). Slope mAG = (0-5)/(3-0) = -5/3. Line EF connects E(5,x) and F(x-5, 0). Slope mEF = (0-x)/((x-5)-5) = -x/(x-10). AG is perpendicular to EF. mAG * mEF = -1. (-5/3) * (-x/(x-10)) = -1 (5x)/(3(x-10)) = -1 5x = -3(x-10) 5x = -3x + 30 8x = 30 x = 30/8 = 15/4. x is CE. So CE = 15/4. All three coordinate system orientations (with appropriate mapping of points) lead to the same result CE=15/4. This makes me confident in the answer. Final confirmation of required output format: structured plain text, covering question stem, options, and chart description with elements and relationships. Question Stem: Done. Options: Done. Chart Description: Done (Type, elements, relationships). Mathematical Formulas/Chemical Equations: Slopes, perpendicular condition (product of slopes = -1). Table Content: Not applicable. Other Relevant Text: Problem source (2020・孝感). The coordinate geometry approach is a valid way to solve this problem and all consistent attempts yielded 15/4. The options confirm that a fractional answer is expected. Let's consider the possibility of error in understanding the diagram or the question. The steps of rotation and perpendicularity are clearly described. The diagram visually supports the relative positions of points and lines. Consider edge cases. If E=D, then DE=0, CE=5. F=B. EF is BD. AG is perpendicular to BD. In a square, the diagonal BD is perpendicular to the diagonal AC. So G must be the intersection of BC and AG. AG being perpendicular to BD might not uniquely determine G on BC. If E=C, then DE=5, CE=0. F is the point obtained by rotating C 90 deg clockwise around A. C=(5,5), A=(0,0). Rotate (5,5) 90 deg clockwise around (0,0) to (5,-5). So F=(5,-5). E=(5,5), F=(5,-5). EF is the vertical line segment from (5,-5) to (5,5). AG is perpendicular to EF, so AG is horizontal. A=(0,0). A horizontal line through (0,0) is y=0, which is the line AB. G is on BC. BC is the line x=5 from y=0 to y=5. The line y=0 intersects the line x=5 at (5,0). So G=(5,0). Then BG is the distance from B(5,0) to G(5,0), which is 0. CG is the distance from C(5,5) to G(5,0), which is 5. This contradicts BG=3, CG=2. So E cannot be C or D. Thus 0 < CE < 5. 15/4 = 3.75, which is between 0 and 5. 5/4 = 1.25, also between 0 and 5. The solution CE=15/4 corresponds to DE=5/4=1.25. E is at (5/4, 5) in the coordinate system A=(0,0), D=(0,5), C=(5,5), B=(5,0). Let's use A=(0,0), B=(5,0), C=(5,5), D=(0,5). E on CD. CD is x-axis from x=0 to x=5, at y=5. E=(xE, 5). DE = xE. CE = 5-xE. Let CE=x. E=(5-x, 5). DE=5-x. Rotate E(5-x, 5) 90 deg clockwise around A(0,0). F=(5, -(5-x))=(5, x-5). G=(5,3). mAG = 3/5. E=(5-x, 5), F=(5, x-5). mEF = ((x-5)-5)/(5-(5-x)) = (x-10)/x. (3/5) * ((x-10)/x) = -1 3(x-10) = -5x 3x - 30 = -5x 8x = 30 x = 30/8 = 15/4. x is CE. CE=15/4. All coordinate systems consistent with the diagram and rotation property lead to the same result. Therefore, the result 15/4 is very likely correct. ``` 1. (2020•孝感)如图,点 E 在正方形 ABCD 的边 CD 上,将△ADE 绕点 A 顺时针旋转 90° 到△ABF 的位置,连接 EF,过点 A 作 EF 的垂线,垂足为点 H,与 BC 交于点 G.若 BG=3, CG=2, 则 CE 的长为 ( ) A. $\frac{5}{4}$ B. $\frac{15}{4}$ C. 4 D. $\frac{9}{2}$ Chart/Diagram Description: Type: Geometric figure. Main Elements: - Shape: A square labeled ABCD. - Points: A, B, C, D are vertices of the square. E is on side CD. F is the image of E after rotating triangle ADE 90 degrees clockwise around A, resulting in triangle ABF. G is on side BC. H is the intersection point of line segment AG and line segment EF, and AG is perpendicular to EF at H. - Lines: Sides AB, BC, CD, DA of the square. Line segments AE, DE, BF, AF, EF, AG. AG is shown intersecting EF at H and BC at G. A right angle symbol is shown at H between AG and EF. - Labels: Points A, B, C, D, E, F, G, H are labeled. - Relative Position and Direction: The square is oriented with A at the top-left, B at the bottom-left, C at the bottom-right, and D at the top-right. AB is the bottom side, BC is the right side, CD is the top side, and DA is the left side. E is shown on the top side CD. G is shown on the right side BC, positioned such that B-G-C order from bottom to top. F is located such that rotating triangle ADE clockwise around A makes it coincide with triangle ABF. H is inside the square, on both AG and EF. ``` The description of the square's orientation based on the diagram is A top-left, B bottom-left, C bottom-right, D top-right. In this case, AB is the left vertical side, BC is the bottom horizontal side, CD is the right vertical side, DA is the top horizontal side. This doesn't match the usual convention or the previous coordinate system attempts. Let's verify the side labels in the diagram. A is top-left, B is bottom-left, C is bottom-right, D is top-right. So AB is vertical, BC is horizontal, CD is vertical, DA is horizontal. This means AB is parallel to CD, and BC is parallel to DA. This doesn't look like a square. Let's assume the vertices are labeled in order around the square. Let's assume the first description of orientation from the text and diagram is correct: A top-left, B bottom-left, C bottom-right, D top-right. Then AB is the left side, BC is the bottom side, CD is the right side, DA is the top side. AB vertical, BC horizontal, CD vertical, DA horizontal. This is still inconsistent. Let's re-read the text: "正方形 ABCD". This means ABCD is a square. The standard notation for a polygon lists vertices in order. So AB, BC, CD, DA are consecutive sides. Let's assume the labels A, B, C, D in the diagram are correct vertices of a square in that order, starting from top-left and going counter-clockwise: A top-left, B bottom-left, C bottom-right, D top-right. Then AB is the left side (vertical), BC is the bottom side (horizontal), CD is the right side (vertical), DA is the top side (horizontal). This forms a square. In this case: A=(0,s), B=(0,0), C=(s,0), D=(s,s). Side length s. G is on BC. BC is on the x-axis from x=0 to x=s. G=(xG, 0). BG = |xG-0| = xG. CG = |xG-s|. BG=3, CG=2. xG=3, s-xG=2. s-3=2, s=5. A=(0,5), B=(0,0), C=(5,0), D=(5,5). This matches the third coordinate system attempt. E is on CD. CD is the vertical segment from (5,5) to (5,0). E=(5, yE). DE is distance from (5, yE) to (5,5) = |yE-5| = 5-yE (since E is on segment CD, 0<=yE<=5). CE is distance from (5, yE) to (5,0) = |yE-0| = yE. Let CE=x. Then E=(5,x), DE=5-x. Rotate E(5,x) 90 deg clockwise around A(0,5). Vector AE = (5-0, x-5) = (5, x-5). Rotate 90 deg clockwise: (x-5, -5). Translate by A(0,5): F = (0+(x-5), 5+(-5)) = (x-5, 0). F is at (x-5, 0). G is at (3,0). Line AG passes through A(0,5) and G(3,0). Slope mAG = (0-5)/(3-0) = -5/3. Line EF connects E(5,x) and F(x-5, 0). Slope mEF = (0-x)/((x-5)-5) = -x/(x-10). mAG * mEF = (-5/3) * (-x/(x-10)) = (5x)/(3(x-10)). (5x)/(3(x-10)) = -1 5x = -3(x-10) = -3x + 30 8x = 30 x = 30/8 = 15/4. CE = x = 15/4. This confirms the result using the coordinate system consistent with the diagram labels as vertices in order and the side lengths from BG and CG. The first coordinate system attempt A=(0,0), B=(5,0), C=(5,5), D=(0,5) assumes AB is horizontal and AD is vertical, starting from A at origin. This doesn't match the diagram orientation but was easy to calculate. However, the rotation calculation in the first attempt had an error initially. The corrected calculation in the first attempt resulted in 15/4. Let's trust the result 15/4. The final output structure is plain text with sections.