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Nt1310 Unit 3 Research Paper

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%\addcontentsline{toc}{section}{Activity 1.14}
\section*{Activity 1.14}

Given ABCD is a square and rhombus EFGH is inscribed in the square. We want to prove that EFGH will be a square.\\
\includegraphics[scale=0.46]{Pictures/actvity1_14_pic.PNG}

\noindent
$EH = HG = GF = FE$\hfill ---\ since $EFGH$ is a rhombus.\\
$AB = BC = DC = DA$\hfill ---\ since $ABCD$ is a square.\\
$BF = DH$\hfill since $AD = BG$ and $EF = GH$ and $EH = GF$.\\
Similarly, $AE = GC$ and $AH = BE = FC = DG$.\\
The four triangles $(AEH;\ EBF;\ FCG;\ DHG)$ are congruent by $SSS$.\\
Each angle in the corner of $ABCD$ is $90^{\circ}$\hfill ---\ since $ABCD$ is a square.\\
$A\hat{E}H + E\hat{H}A = 90^{\circ}$\hfill ---\ since $E\hat{A}H = 90^{\circ}$.\\
Similarly $B\hat{E}F …show more content…

I only prove that $E\hat{H}G = 90^{\circ}$, for the reason that the prove of the other angles will be similar. When I was asked to prove if the inscribed rhombus is a square, I simply say to myself, ``it means I have to prove that the corners of the rhombus are right angled ($90^{\circ}$)" since the properties of a rhombus are such that opposite angles are equal, all sides are equal, and opposite sides are parallel, the diagonals bisect the angles, and the diagonals are perpendicular bisectors of each other. Therefore, the inscribed rhombus can only be a square if the angles are right angles. I observe that all four triangles are congruent by SSS, therefore from $A\hat{E}H + E\hat{H}A = 90^{\circ}$ and $E\hat{H}G + A\hat{H}E + D\hat{H}G = 180^{\circ}$ and $G\hat{H}D = A\hat{E}H$, it follows that $A\hat{H}E + D\hat{H}G = 90^{\circ}$, hence $E\hat{H}G = 90^{\circ}$. There is a need to draw the picture since it gives more logical reasoning and the understanding of how the angles and sides are defined, without drawing the picture we may not understand what is happening, therefore it is very crucial that as teachers we make use of pictures when we dealing with this kind of

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