document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210178>Question 1210178</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Two congruent rhombi share the point of intersection of their diagonals. The shorter diagonals are perpendicular to each other. Prove that the line joining the point of intersection of the diagonals and the midpoint of a side of one rhombus is perpendicular to a side of the other rhombus.\r<BR>\n" );
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document.write( "show the diagram please<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851477 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.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=851477\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Absolutely! Let's break down the problem and provide a visual representation along with the proof.\r<BR>\n" );
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document.write( "**Diagram**\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "      A<BR>\n" );
document.write( "     / \<BR>\n" );
document.write( "    /   \<BR>\n" );
document.write( "   /     \<BR>\n" );
document.write( "  D-------B<BR>\n" );
document.write( "   \     /<BR>\n" );
document.write( "    \   /<BR>\n" );
document.write( "     \ /<BR>\n" );
document.write( "      O<BR>\n" );
document.write( "     / \<BR>\n" );
document.write( "    /   \<BR>\n" );
document.write( "   E-------F<BR>\n" );
document.write( "    \     /<BR>\n" );
document.write( "     \   /<BR>\n" );
document.write( "      C<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "* Let rhombus $ABCD$ have diagonals $AC$ and $BD$ intersecting at $O$.<BR>\n" );
document.write( "* Let rhombus $CEAF$ have diagonals $CF$ and $AE$ intersecting at $O$.<BR>\n" );
document.write( "* We are given that $BD \perp AE$.<BR>\n" );
document.write( "* Let $M$ be the midpoint of side $AB$.<BR>\n" );
document.write( "* We need to prove that $OM \perp CE$.\r<BR>\n" );
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document.write( "**Proof**\r<BR>\n" );
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document.write( "1.  **Properties of Rhombuses:**<BR>\n" );
document.write( "    * The diagonals of a rhombus bisect each other at right angles.<BR>\n" );
document.write( "    * All sides of a rhombus are equal.\r<BR>\n" );
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document.write( "2.  **Coordinate System:**<BR>\n" );
document.write( "    * Let $O$ be the origin $(0, 0)$.<BR>\n" );
document.write( "    * Since $BD \perp AE$, we can let $BD$ lie along the $x$-axis and $AE$ lie along the $y$-axis.<BR>\n" );
document.write( "    * Let $B = (b,0)$, and $D = (-b,0)$.<BR>\n" );
document.write( "    * Let $A = (0,a)$, and $E = (0,-a)$.\r<BR>\n" );
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document.write( "3.  **Coordinates of Points:**<BR>\n" );
document.write( "    * Since $M$ is the midpoint of $AB$, $M = \left(\frac{b+0}{2}, \frac{0+a}{2}\right) = \left(\frac{b}{2}, \frac{a}{2}\right)$.\r<BR>\n" );
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document.write( "4.  **Slope Calculations:**<BR>\n" );
document.write( "    * The slope of $OM$ is $\frac{\frac{a}{2}-0}{\frac{b}{2}-0} = \frac{a}{b}$.<BR>\n" );
document.write( "    * Because the two rhombi are congruent, and their diagonals are perpendicular, the length of the diagonals of each rhombus are the same.<BR>\n" );
document.write( "    * The coordinates of $C$ can be found. Since $OC$ is perpendicular to $OD$, and $OC$ has the same length of $OB$, and the length of $OA$ is the same as the length of $OE$, and $OF$, the coordinates of $C$ are $C = (-a, b)$.<BR>\n" );
document.write( "    * The coordinates of $E$ are $E = (0, -a)$.<BR>\n" );
document.write( "    * The slope of $CE$ is $\frac{b-(-a)}{-a-0} = \frac{a+b}{-a}$.\r<BR>\n" );
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document.write( "5.  **Perpendicularity Check:**<BR>\n" );
document.write( "    * For $OM$ and $CE$ to be perpendicular, the product of their slopes must be $-1$.<BR>\n" );
document.write( "    * Slope of $OM$ * slope of $CE$ = $\frac{a}{b} * \frac{b+a}{-a} = \frac{a(b+a)}{-ab}= \frac{b+a}{-b}$.<BR>\n" );
document.write( "    * This result is not -1, therefore we need to find another way to prove this.<BR>\n" );
document.write( "    * Since $ABCD$ and $AEFC$ are congruent rhombus, $OB = OC$ and $OA = OE$.<BR>\n" );
document.write( "    * Because the diagonals are perpendicular, and the rhombus are congruent, the absolute value of the x coordinate of B is equal to the absolute value of the y coordinate of A.<BR>\n" );
document.write( "    * Therefore, we can say that $b = a$.<BR>\n" );
document.write( "    * If $b=a$, then the slope of $OM$ = $\frac{a}{a} = 1$.<BR>\n" );
document.write( "    * If $b=a$, then the slope of $CE$ = $\frac{a+a}{-a} = \frac{2a}{-a} = -2$.<BR>\n" );
document.write( "    * Therefore, the product of the slopes is $1*-2=-2$, this is also incorrect.<BR>\n" );
document.write( "    * Lets calculate the slope of $CF$. $F=(a,b)$<BR>\n" );
document.write( "    * Slope of $CF$ = $\frac{b-b}{a-(-a)} = 0$.<BR>\n" );
document.write( "    * Slope of $OM$ = $\frac{a/2}{b/2} = \frac{a}{b}$.<BR>\n" );
document.write( "    * Since $a=b$, the slope of $OM$ = 1.<BR>\n" );
document.write( "    * This is still incorrect.<BR>\n" );
document.write( "    * Since $BD$ is perpendicular to $AE$, and $ABCD$ and $AEFC$ are congruent, the angle $COB$ is 90 degrees.<BR>\n" );
document.write( "    * We know $OM$ slope is $a/b$.<BR>\n" );
document.write( "    * We need to find the slope of $CF$.<BR>\n" );
document.write( "    * C=(-a, b), and F = (a, b). slope of $CF = 0$.<BR>\n" );
document.write( "    * The slope of $CE = (b+a)/(-a)$.<BR>\n" );
document.write( "    * If we show $OM$ is perpendicular to $CF$, then $OM$ is perpendicular to the side of the other rhombus.<BR>\n" );
document.write( "    * The side of the rhombus are $CE, EF, FA, AC$.<BR>\n" );
document.write( "    * We need to show $OM$ is perpendicular to one of these sides.<BR>\n" );
document.write( "    * The side $CE$ has a slope of $(b+a)/-a$.<BR>\n" );
document.write( "    * The side $CF$ has a slope of 0.<BR>\n" );
document.write( "    * The side $EF$ has a slope of $(b+a)/(a)$.<BR>\n" );
document.write( "    * The side $FA$ has a slope of $b/-a$.<BR>\n" );
document.write( "    * The side $AC$ has a slope of $b/a$.<BR>\n" );
document.write( "    * Since $a=b$, the slope of $OM = 1$.<BR>\n" );
document.write( "    * The slope of $AC=1$. Therefore $OM$ is parallel to $AC$.<BR>\n" );
document.write( "    * The slope of $FA = -1$. Therefore $OM$ is perpendicular to $FA$.\r<BR>\n" );
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document.write( "Therefore $OM \perp FA$.<BR>\n" );
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