Question 1165628: Given: AB= DC
prove: AC= DB
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2103)   (Show Source):
You can put this solution on YOUR website! This is a great question that tests the definitions and properties of quadrilaterals! However, based on the standard geometric definitions, the statement you are asked to prove, **AC = DB**, is **not generally true** for a kite with the given condition.
In standard Euclidean geometry:
1. A **Kite** is a quadrilateral where two pairs of equal-length sides are adjacent to each other.
2. The diagonals of a kite are **perpendicular** ($\overline{AC} \perp \overline{DB}$).
3. Only **one** of the diagonals is bisected (the shorter one).
The statement **AC = DB** means the diagonals are equal, which is a property of a **rectangle** or an **isosceles trapezoid**, but not a general kite.
Let's assume there might be a property of the specific kite *ABCD* implied by the side equality $\mathbf{AB = DC}$.
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## 📐 Analysis of the Given Information
### Given: ABCD is a Kite
The standard side properties of a kite are that two pairs of adjacent sides are equal. There are two possible pairings:
1. **Case 1 (Standard):** $AB = AD$ and $CB = CD$.
2. **Case 2 (Alternative):** $AB = CB$ and $AD = CD$.
### Given: AB = DC
The given condition **$AB = DC$** (adjacent sides equal to non-adjacent sides) is unusual for a general kite, but we must incorporate it.
If $ABCD$ is a kite, we must combine the kite definition with the given condition:
* **If we assume Case 1 ($AB = AD$ and $CB = CD$):**
* We are given $AB = DC$.
* Since $AB=AD$ and $CB=CD$, substituting the given condition means $AD = DC$.
* Since $AD = DC$ and $CB = CD$, then $AD=DC=CB$.
* If all three sides ($AD, DC, CB$) are equal, the only way $ABCD$ can be a kite is if the fourth side is also equal, making **$ABCD$ a Rhombus** (and thus a Parallelogram).
* **If $ABCD$ is a Rhombus:** All sides are equal ($AB=BC=CD=DA$). A rhombus is a specific type of kite.
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## 🛑 Proving the Diagonals are Equal
The statement you want to prove, $\mathbf{AC = DB}$, is only true if the figure is a **rectangle** (or a square).
If a quadrilateral is *both* a kite (diagonals are perpendicular) *and* a rectangle (diagonals are equal), it must be a **Square**.
### Conclusion based on the Given Information
If the given conditions **$AB=DC$** and **$ABCD$ is a kite** force the figure to be a **Rhombus** (as shown above), then:
* Rhombus properties: Diagonals are perpendicular.
* **Rhombus properties: Diagonals are NOT necessarily equal.**
**Conclusion:** The claim $\mathbf{AC = DB}$ cannot be proven from the given information alone.
If this were intended to be a textbook problem that *is* provable, the quadrilateral must have been defined as something else, such as an **Isosceles Trapezoid** (where $\overline{AC} = \overline{DB}$), or the side condition was meant to imply the figure is a **Square** (which would be an extremely specific case).
**The statement AC = DB is false for the general conditions AB=DC and ABCD is a kite.**
Answer by ikleyn(53250)  (Show Source):
You can put this solution on YOUR website! .
Given: AB= DC
prove: AC= DB
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The problem formulation in the post is FATALLY INCOMPLETE,
and therefore the problem can not be solved properly.
The logic and the reasoning in the post by @CPhill are irrelevant
to the possible case. @CPhill starts from consideration of a kite shape,
which is 100% not relevant to the given equalities.
If this relates to quadrilaterals on the plane, then only two cases
can be relevant: (a) parallelograms, and (b) isosceles trapezoids.
But then the problem's description must provide relevant conditions
for such a quadrilateral to be a parallelogram or an isosceles trapezoid.
Without it, the problem is the same absurdist non-sensical creation
as a two-leg horse from the story about the baron Munchausen
and can exist only in sick imagination or in tales for children to make them laugh.
Posting such incomplete defective "problems" to this forum is unacceptable hooliganism.
The way how @CPhill reacts to such absurdist problems, is a hooliganism in degree 7 (seven).
He piles nonsense on top of other nonsense.
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