SOLUTION: In rectangle ABCD, corner A is folded over crease DE to point F on BC. Find BC.

Question 1210474: In rectangle ABCD, corner A is folded over crease DE to point F on BC. Find BC.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2136) (Show Source): You can put this solution on YOUR website!
The problem as stated is **unsolvable** because a necessary dimension is missing. The length of the side $BC$ (which is equal to $AD$) cannot be uniquely determined without knowing the length of the adjacent side **$CD$** (or $AB$).
Here is the general geometric relationship that must hold in this type of folding problem, which shows the dependency:
Let:
* $BC = AD = l$ (The length you are trying to find)
* $CD = w$ (The missing width)
### 1. Relationships from the Fold
When corner $A$ is folded over the crease $DE$ to point $F$ on $BC$, two triangles, $\triangle ADE$ and $\triangle FDE$, are congruent. This gives us the following equivalences:
* The hypotenuse of the fold: $DF = AD = l$
* The fold sides: $AE = FE$
* $\angle DFE = \angle DAE = 90^\circ$
### 2. Using the Pythagorean Theorem
We can set up two equations using the Pythagorean theorem in the right-angled triangles $\triangle FDC$ and $\triangle FCE$:
* **In $\triangle FDC$ (Right-angled at C):**
$$DF^2 = DC^2 + CF^2$$
$$l^2 = w^2 + CF^2 \quad \rightarrow \quad CF = \sqrt{l^2 - w^2}$$
* **In $\triangle FCE$ (Right-angled at C):**
$$FE^2 = CE^2 + CF^2$$
Since $FE = AE$, we substitute $AE$ and the expression for $CF^2$:
$$AE^2 = CE^2 + (l^2 - w^2)$$
### 3. Using the Side $CD$
The line segments $DE$ and $CE$ make up the side $CD$: $CD = DE + CE$.
* **In $\triangle ADE$ (Right-angled at D):**
$$AE^2 = AD^2 + DE^2$$
$$AE^2 = l^2 + DE^2$$
### Conclusion of Dependency
By setting the two expressions for $AE^2$ equal to each other:
$$CE^2 + l^2 - w^2 = l^2 + DE^2$$
$$\rightarrow CE^2 = DE^2 + w^2$$
Since $w$ and $l$ are not given, $BC$ (which is $l$) cannot be uniquely determined.
**To solve the problem, you must know the length of $CD$ (or a relationship between $l$ and $w$).**
Answer by ikleyn(53266) (Show Source): You can put this solution on YOUR website!
.
In rectangle ABCD, corner A is folded over crease DE to point F on BC. Find BC.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Make a sketch following my descriptions/instructions. We consider rectangle ABCD. For simplicity, assume that the rectangle is "vertical", i.e. its vertical dimension 'b' = AD = BC is greater than (or equal to) its horizontal dimension 'a' = AB = CD. We fold corner A, so point A comes to point F on BC: so, point F is the image of point A under folding, and F is on BC between points B and C. <<<---=== It is in accordance with the given part. Then it is clear that at this folding the side AB transforms to interval BF; so, AB = BF = a. It is also clear that at this folding the side AD becomes FD; therefore, DF = b. Consider right-angled triangle FCD. For its leg FC we can write FC = = . Now interval BC is the sum of two intervals BC = BF + FC = a + . But BC = b; so, we can rewrite the last equation in the form b = a + . (1) Now we will perform formal transformations with this equation. From (1) we have, moving 'a' on the left side b - a = . Square both sides (b-a)^2 = b^2 - a^2, b^2 - 2ab + a^2 = b^2 - a^2 Cancel b^2 in both sides and continue -2ab + a^2 = - a^2, a^2 + a^2 = 2ab, 2a^2 = 2ab. Cancel the common factor 2a in both sides a = b. So, we proved that under given conditions, it must be a = b. It is what the problem wants to be proved. ANSWER. Under given conditions, the rectangle ABCD is the square and its sides are congruent: a = b.

Solved.

It is what the Artificial Intelligence (= @CPhill) missed and could not to prove properly.

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