document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210498>Question 1210498</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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 $ABCD$. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852995 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852995\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Based on the given information, the area of the parallelogram $ABCD$ is $\mathbf{8}$.\r<BR>\n" );
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document.write( "Here is the step-by-step reasoning.\r<BR>\n" );
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document.write( "## 📐 Analysis and Calculation\r<BR>\n" );
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document.write( "### Step 1: Analyze the Area of Triangle $ABC$\r<BR>\n" );
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document.write( "In a parallelogram $ABCD$, the diagonal $\overline{AC}$ divides the parallelogram into two congruent triangles: $\triangle ABC$ and $\triangle CDA$.<BR>\n" );
document.write( "Therefore, their areas are equal:<BR>\n" );
document.write( "$$Area(\triangle CDA) = Area(\triangle ABC)$$<BR>\n" );
document.write( "We are given that $Area(\triangle ABC) = 4$.<BR>\n" );
document.write( "$$Area(\triangle CDA) = 4$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### Step 2: Relate the Area of $\triangle ABC$ to the Area of Parallelogram $ABCD$\r<BR>\n" );
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document.write( "The area of the parallelogram $ABCD$ is the sum of the areas of the two congruent triangles:<BR>\n" );
document.write( "$$Area(ABCD) = Area(\triangle ABC) + Area(\triangle CDA)$$<BR>\n" );
document.write( "$$Area(ABCD) = 4 + 4$$<BR>\n" );
document.write( "$$\mathbf{Area(ABCD) = 8}$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### Step 3: Analyze the Additional Information\r<BR>\n" );
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document.write( "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 $Area(\triangle ABC)$.\r<BR>\n" );
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document.write( "However, this redundant information is useful for **locating point $E$**.\r<BR>\n" );
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document.write( "* $\triangle EBC$ and $\triangle ABC$ share the same base $\overline{BC}$.<BR>\n" );
document.write( "* The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.<BR>\n" );
document.write( "    $$Area(\triangle EBC) = \frac{1}{2} \times BC \times h_E$$<BR>\n" );
document.write( "    $$Area(\triangle ABC) = \frac{1}{2} \times BC \times h_A$$<BR>\n" );
document.write( "    Where $h_E$ is the perpendicular distance from $E$ to $\overline{BC}$, and $h_A$ is the perpendicular distance from $A$ to $\overline{BC}$.\r<BR>\n" );
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document.write( "Since $Area(\triangle EBC) = 4$ and $Area(\triangle ABC) = 4$, we have:<BR>\n" );
document.write( "$$\frac{1}{2} \times BC \times h_E = \frac{1}{2} \times BC \times h_A$$<BR>\n" );
document.write( "$$h_E = h_A$$\r<BR>\n" );
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document.write( "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 $A$.\r<BR>\n" );
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document.write( "* If $\mathbf{E = A}$, then $Area(\triangle EBC) = Area(\triangle ABC) = 4$.<BR>\n" );
document.write( "* 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.<BR>\n" );
document.write( "* However, if lines $DA$ and $BC$ are parallel, they never intersect, meaning $F$ would not exist unless we consider $F$ to be at infinity.\r<BR>\n" );
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document.write( "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.\r<BR>\n" );
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document.write( "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.\r<BR>\n" );
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document.write( "$$Area(ABCD) = 2 \times Area(\triangle ABC) = 2 \times 4 = 8$$\r<BR>\n" );
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document.write( " </SPAN>\n" );
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