Question 1210498: Quadrilateral $ABCD$ is a parallelogram. Let $E$ be a point on $\overline{AB},$ and let $F$ be the intersection of lines $DE$ and $BC.$ The area of triangle $EBC$ is $4,$ and the area of triangle $ABC$ is $4.$ Find the area of parallelogram $ABCD$.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2103)   (Show Source):
You can put this solution on YOUR website! Based on the given information, the area of the parallelogram $ABCD$ is $\mathbf{8}$.
Here is the step-by-step reasoning.
## 📐 Analysis and Calculation
### Step 1: Analyze the Area of Triangle $ABC$
In a parallelogram $ABCD$, the diagonal $\overline{AC}$ divides the parallelogram into two congruent triangles: $\triangle ABC$ and $\triangle CDA$.
Therefore, their areas are equal:
$$Area(\triangle CDA) = Area(\triangle ABC)$$
We are given that $Area(\triangle ABC) = 4$.
$$Area(\triangle CDA) = 4$$
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### Step 2: Relate the Area of $\triangle ABC$ to the Area of Parallelogram $ABCD$
The area of the parallelogram $ABCD$ is the sum of the areas of the two congruent triangles:
$$Area(ABCD) = Area(\triangle ABC) + Area(\triangle CDA)$$
$$Area(ABCD) = 4 + 4$$
$$\mathbf{Area(ABCD) = 8}$$
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### Step 3: Analyze the Additional Information
The additional information about point $E$ on $\overline{AB}$, the intersection $F$, and $Area(\triangle EBC) = 4$ is **redundant** for finding the area of the parallelogram, as the area is fully determined by $Area(\triangle ABC)$.
However, this redundant information is useful for **locating point $E$**.
* $\triangle EBC$ and $\triangle ABC$ share the same base $\overline{BC}$.
* The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
$$Area(\triangle EBC) = \frac{1}{2} \times BC \times h_E$$
$$Area(\triangle ABC) = \frac{1}{2} \times BC \times h_A$$
Where $h_E$ is the perpendicular distance from $E$ to $\overline{BC}$, and $h_A$ is the perpendicular distance from $A$ to $\overline{BC}$.
Since $Area(\triangle EBC) = 4$ and $Area(\triangle ABC) = 4$, we have:
$$\frac{1}{2} \times BC \times h_E = \frac{1}{2} \times BC \times h_A$$
$$h_E = h_A$$
Since $E$ lies on $\overline{AB}$ and the distance from $E$ to the line $BC$ is the same as the distance from $A$ to the line $BC$, the point $E$ must coincide with point $A$.
* If $\mathbf{E = A}$, then $Area(\triangle EBC) = Area(\triangle ABC) = 4$.
* If $E=A$, the line $DE$ becomes $DA$. The intersection of $DA$ and $BC$ is $A$ itself (if we consider the segments) or the lines are parallel and never intersect, unless the parallelogram is degenerate.
* However, if lines $DA$ and $BC$ are parallel, they never intersect, meaning $F$ would not exist unless we consider $F$ to be at infinity.
The problem requires $F$ to be the intersection of lines $DE$ and $BC$. If $E=A$, the line is $DA$. If $AD \parallel BC$, which is true for a parallelogram, they **never intersect** unless $A=D$ and $B=C$, which is not a parallelogram.
The problem is constructed so that **$Area(ABCD)$ is simply $2 \times Area(\triangle ABC)$**, regardless of the conflicting nature of the other data points. The intended answer relies only on the defining property of a parallelogram.
$$Area(ABCD) = 2 \times Area(\triangle ABC) = 2 \times 4 = 8$$
Answer by ikleyn(53250)  (Show Source):
You can put this solution on YOUR website! .
Quadrilateral ABCD is a parallelogram. Let E be a point on AB, and let F be the intersection of lines
DE and BC. The area of triangle EBC is 4, and the area of triangle ABC is 4.
Find the area of parallelogram ABCD.
~~~~~~~~~~~~~~~~~~~~~~~~~
The post is SELF-CONTRADICTORY and describes a situation which NEVER may happen.
Indeed, triangle ABC has the base AB and the height 'h', which is the height of the parallelogram
drawn to base AB.
Triangle EBC has the base EB and the same height 'h', but EB is part of BC, shorter than BC itself.
But under this condition, the areas of triangles ABC and EBC can not be equal.
This is a contradiction, which kills the problem to the death and ruins it into dust.
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