Resources for Proving Properties of Quadrilaterals
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Questions
5
With Worked SolutionClick Here -
Video Tutorials
1
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Proving Properties of Quadrilaterals Theory
![\begin{multicols}{2} \textbf{Example 1}\\ \(A B C D\) is a trapezium \(A D=B C,\; A E=B F\) \\ and \(\angle A E D=\angle B F C=90^{\circ}\)\\ Prove that \(\angle A D E=\angle B C F\)\\ \begin{center} \begin{tikzpicture}[scale=1.2] \coordinate[label=below left:D] (O) at (0,0); \coordinate[label=above left:A] (A) at (70:2cm); \begin{scope}[shift=(A)] \coordinate[label=above right:B] (B) at (right:2cm); \end{scope} \begin{scope}[shift=(B)] \coordinate[label=below right:C] (C) at (290:2cm); \end{scope} \coordinate[label=below:E] (D) at ($(O)!(A)!(C)$); \coordinate[label=below:F] (E) at ($(O)!(B)!(C)$); \draw[line width=1pt] (O)--(A) node[midway,sloped]{\(|\)}--(B)node[midway]{>}--(C)node[midway,sloped]{\(|\)}--(O)node[midway]{>}; \draw[dashed,line width=1pt] (A)--(D); \draw[dashed,line width=1pt] (B)--(E); \pic[line width=1pt, draw, angle radius=3mm] {right angle= A--D--O}; \pic[line width=1pt, draw, angle radius=3mm] {right angle= B--E--C}; \end{tikzpicture} \end{center} \textbf{Example 1 solution}\\ In \(\triangle's \; AED,\; BFC\)\\ \(\begin{aligned} A D&=B C &&(\text{ given })\\ AE&=BF &&(\text{ given })\\ \angle AED&=\angle B F C=90^{\circ} &&(\text{ given })\\ \therefore \triangle A E D &\equiv \triangle BFC && (\text{ RHS })\\ \therefore \angle A D E&=\angle BCF \end{aligned}\)\\ (corresponding \(\angle^{\prime}\) in congruent \(\left.\triangle's\right)\) \columnbreak \textbf{Example 2}\\ \(A B C D\) is a parallelogram\\ Prove that \(\angle A D C=\angle A B C\)\\ \begin{center} \begin{tikzpicture}[scale=1] \coordinate[label=below left:D] (O) at (0,0); \coordinate[label=above left:A] (A) at (60:1.5cm); \coordinate[label=below right:C] (B) at (right:3cm); \begin{scope}[shift=(B)] \coordinate[label=above right:B] (C) at (60:1.5cm); \end{scope} \draw[line width=1pt] (O)--(B)node[midway,rotate=90]{}--(C)node[midway]{ }--(A)node[midway,rotate=90]{}--(O)node[midway]{}--cycle; \path (O)--(B) node[midway,sloped] {\(>\)}; \path (A)--(C) node[midway,sloped] {\(>\)}; \path (O)--(A) node[pos=0.7,sloped] {\(>\)}; \path (B)--(C) node[pos=0.7,sloped] {\(>\)}; \path (O)--(A) node[pos=0.8,sloped] {\(>\)}; \path (B)--(C) node[pos=0.8,sloped] {\(>\)}; \draw[line width=1pt,densely dashed] (A)--(B); \end{tikzpicture} \end{center} \textbf{Example 2 solution}\\ \(A B C D\) is a parallelogram\\ Prove that \(\angle A D C=\angle A B C\)\\ In \(\triangle\)'s \(ADC,\; ABC\)\\ \(\begin{aligned} A D&=C B &&(\text{opposite sides of a parallelogram }=)\\ AB&=CD &&(\text{opposite sides of a parallelogram} =) \end{aligned}\)\\ \(AC\) is common\\ \(\therefore \triangle A D C \equiv \triangle A B C \text { (SSS) }\)\\ \(\therefore \angle A D C=\angle A B C\)\\ (corresponding \(\angle\) 's of congruent \(\triangle^{\prime} s\) ) \end{multicols}](/media/i2hgvtp3/32395.png)