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如图，在菱形纸片 ABCDA B C D 中， ∠A=60∘\angle \mathrm{A}=60^{\circ} ，将纸片折刍，点 A、DA 、 D 分别落在 $A^{\...
Mar 9, 2024
如图，在菱形纸片 ABCDA B C D 中， ∠A=60∘\angle \mathrm{A}=60^{\circ} ，将纸片折刍，点 A、DA 、 D 分别落在 A′、DA^{\prime} 、 D 处，且 A′DA^{\prime} D 经过 B，EFB ， E F 为折痕，当 D′F⊥CDD^{\prime} F \perp C D 时， CFFD\frac{C F}{F D} 的值为
Solution by Steps
step 1
Given a rhombus ABCD ABCD with ∠A=60∘ \angle A = 60^\circ , when the paper is folded such that point A A and D D fall on A′ A' and D′ D' respectively, and A′D A'D passes through B B , with EF EF being the crease, and D′F⊥CD D'F \perp CD , we need to find the value of CFFD \frac{CF}{FD}
step 2
Since ABCD ABCD is a rhombus, all sides are equal. Let the side of the rhombus be s s
step 3
Triangle ABD ABD is an equilateral triangle because ∠A=60∘ \angle A = 60^\circ and AB=AD=BD=s AB = AD = BD = s
step 4
When the paper is folded, D′F⊥CD D'F \perp CD and D′F D'F bisects ∠D \angle D , creating two 30∘−60∘−90∘ 30^\circ-60^\circ-90^\circ triangles, D′FC D'FC and D′FD D'FD
step 5
In a 30∘−60∘−90∘ 30^\circ-60^\circ-90^\circ triangle, the sides are in the ratio 1:3:2 1:\sqrt{3}:2 . Therefore, FD=s2 FD = \frac{s}{2} and CD=s3 CD = s\sqrt{3}
step 6
Since CD=CF+FD CD = CF + FD , we have s3=CF+s2 s\sqrt{3} = CF + \frac{s}{2}
step 7
Solving for CF CF , we get CF=s3−s2 CF = s\sqrt{3} - \frac{s}{2}
step 8
To find CFFD \frac{CF}{FD} , we substitute the values of CF CF and FD FD into the ratio
step 9
Thus, CFFD=s3−s2s2 \frac{CF}{FD} = \frac{s\sqrt{3} - \frac{s}{2}}{\frac{s}{2}}
step 10
Simplifying the ratio, we get CFFD=23−1 \frac{CF}{FD} = 2\sqrt{3} - 1
Answer
CFFD=23−1 \frac{CF}{FD} = 2\sqrt{3} - 1
Key Concept
Properties of rhombus and special right triangles
Explanation
In a rhombus, all sides are equal, and the diagonals bisect the angles. The 30∘−60∘−90∘ 30^\circ-60^\circ-90^\circ triangle has sides in a fixed ratio, which allows us to find the lengths of the segments and the ratio CFFD \frac{CF}{FD} .
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