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Congruent Triangle Proof Theory
![\begin{multicols}{2} \textbf{Example 1}\\ In \(\triangle A B C\), $$ \begin{aligned} & A B=A C \\ & A D \perp B C \\ & B D=D C \end{aligned} $$ Prove that \(\triangle A B D \equiv \triangle A C D\)\\ \begin{center} \begin{tikzpicture}[scale=1] \coordinate[label=below:D] (O) at (0,0); \coordinate[label=below right:C] (A) at (1.5,0); \coordinate[label=below left:B] (B) at (-1.5,0); \coordinate[label=above:A] (C) at (0,3); \draw[line width=1pt] (O)--(A) node[sloped,midway]{\(|\)}--(C)node[sloped,midway]{\(||\)}--(B)node[sloped,midway]{\(||\)}--(O)node[sloped,midway]{\(|\)}; \draw[line width=1pt,rounded corners=1pt] (O)--(C); \pic [draw=black,line width=1pt,angle radius=0.35cm,angle eccentricity=1.6,""] {right angle=C--O--B}; \end{tikzpicture} \end{center} \textbf{Example 1 solution}\\ \(\begin{aligned} AB&=A C &&(\text { given }) \\ \angle A D B&=\angle A D C=90^{\circ} &&(A D \perp B C) \\ B D&=C D &&(\text { given }) \\ \therefore \triangle A B D &\equiv \triangle A C D &&(R H S) \end{aligned}\) \columnbreak \textbf{Example 2}\\ Prove that \(\triangle P R Q \equiv \triangle T R S\)\\ \begin{center} \begin{tikzpicture}[scale=1.2] \coordinate[label=below:R] (O) at (0,0); \coordinate[label=above right:S] (A) at (1,1); \coordinate[label=above left:P] (B) at (-1,1); \coordinate[label=below right:T] (C) at (1,-1); \coordinate[label=below left:Q] (D) at (-1,-1); \draw[line width=1pt] (O)--(A) node[sloped,midway]{\(||\)}--(C)node[sloped,midway]{}--(O)node[sloped,midway]{\(|\)}--(D)node[sloped,midway]{\(||\)}--(B)node[sloped,midway]{}--(O)node[sloped,midway]{\(|\)}; \end{tikzpicture} \end{center} \textbf{Example 2 solution}\\ \(\begin{aligned} PR&=TR &&(\text { given }) \\ QR&=SR && (\text { given }) \\ \angle PRQ&=\angle TRS &&(\text { vertically opposite } \angle's\;) \\ \therefore \triangle PRQ &\equiv \triangle TRS && (\text{SAS}) \end{aligned}\) \end{multicols}](/media/okhlmsnb/32394.png)
NSW Y8 Maths Congruent Figures Congruent Triangle Proof
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