\displaystyle \textbf{Chapter: Inequalities (Triangles)}

\displaystyle \textbf{INTRODUCTION}
\displaystyle \text{1. The sign }>\text{ means, "is greater than", i.e. if }a\text{ is greater than }b, \\ \text{ we write }a>b.
\displaystyle \text{2. The sign }<\text{ means, "is less than", i.e. if }a\text{ is less than }b, \\ \text{ we write }a<b.

\displaystyle \\

\displaystyle \textbf{Theorem 3}
\displaystyle \textit{\text{If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.}}
\displaystyle \textbf{Given: }\triangle ABC\text{ in which }AB>AC.
\displaystyle \textbf{To Prove: }\angle ACB>\angle B.
\displaystyle \textbf{Construction: }\text{From }AB,\text{ cut }AD=AC.\text{ Join }C\text{ and }D.\displaystyle \textbf{Proof:}
\displaystyle \textbf{Statement:}\hspace{2cm}\textbf{Reason:}
\displaystyle \text{In }\triangle ACD:
\displaystyle \text{1. }AC=AD\hspace{2.7cm}\text{By construction}
\displaystyle \text{2. }\angle ACD=\angle ADC\hspace{1cm}\text{Angles opposite to equal sides}
\displaystyle \text{In }\triangle BDC:
\displaystyle \text{3. Ext. }\angle ADC>\angle B\hspace{0.8cm}\text{Exterior angle of a triangle is always}
\displaystyle \hspace{5.8cm}\text{greater than each of its interior opposite angles.}
\displaystyle \therefore\ \angle ACD>\angle B\hspace{1.8cm}\text{From 2 and 3.}
\displaystyle \therefore\ \angle ACB>\angle B\hspace{1.4cm}\text{Since }\angle ACD\text{ is a part of }\angle ACB,
\displaystyle \hspace{5.8cm}\therefore\ \angle ACB>\angle ACD>\angle B.
\displaystyle \therefore\ \textbf{Hence Proved.}

\displaystyle \\

\displaystyle \textbf{Theorem 4 (Converse of Theorem 3)}
\displaystyle \textit{\text{If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.}}
\displaystyle \textbf{Given: }\triangle ABC\text{ in which }\angle CAB>\angle B.
\displaystyle \textbf{To Prove: }BC>AC.
\displaystyle \textbf{Construction: Draw }\angle BAD=\angle B.\displaystyle \textbf{Proof:}
\displaystyle \textbf{Statement:}\hspace{2cm}\textbf{Reason:}
\displaystyle \text{In }\triangle ABD:
\displaystyle \text{1. }AD=BD\hspace{2.7cm}\text{Sides opposite to equal angles.}
\displaystyle \text{In }\triangle ADC:
\displaystyle \text{2. }AD+DC>AC\hspace{1.6cm}\text{Sum of any two sides of a triangle is}
\displaystyle \hspace{5.8cm}\text{always greater than the third side.}
\displaystyle \therefore\ BD+DC>AC\hspace{1.5cm}\text{Since }AD=BD.
\displaystyle \therefore\ BC>AC\hspace{2.8cm}\text{Since }BD+DC=BC.
\displaystyle \therefore\ \textbf{Hence Proved.}

\displaystyle \\

\displaystyle \textbf{Theorem 5}
\displaystyle \textit{\text{Of all the lines that can be drawn to a given straight line from a given point}}
\displaystyle \textit{\text{outside it, the perpendicular is the shortest.}}
\displaystyle \textbf{Given: }\text{A point }O\text{ outside the line }AB\text{ and }OP\perp AB.
\displaystyle \textbf{To Prove: }OP\text{ is the shortest of all the lines that can be drawn from }O\text{ to }AB.
\displaystyle \textbf{Construction: Join }O\text{ with any point }Q\text{ on }AB.\displaystyle \textbf{Proof:}
\displaystyle \textbf{Statement:}\hspace{2cm}\textbf{Reason:}
\displaystyle \text{In right-angled }\triangle OPQ:
\displaystyle \text{1. }\angle OPQ>\angle OQP\hspace{1cm}\text{Right angle is the greatest angle in a} \\ \text{right-angled triangle.}
\displaystyle \text{2. }OQ>OP\hspace{2.5cm}\text{Side opposite to greater angle is greater.}
\displaystyle \text{Similarly, it can be shown that }OP\text{ is smaller than any other line that can be drawn} \\ \text{from }O\text{ to }AB.
\displaystyle \therefore\ OP\text{ is the shortest line drawn from }O\text{ to }AB.
\displaystyle \therefore\ \textbf{Hence Proved.}

\displaystyle \\

\displaystyle \textbf{Corollary 1: }\text{The sum of the lengths of any two sides of a triangle is always greater} \\ \text{than the third side.}
\displaystyle \textbf{For example:}\displaystyle \text{In }\triangle ABC:
\displaystyle \text{(i) }AB+AC>BC,
\displaystyle \text{(ii) }AB+BC>AC\text{ and}
\displaystyle \text{(iii) }BC+AC>AB.

\displaystyle \\

\displaystyle \textbf{Corollary 2: }\text{The difference between the lengths of any two sides of a triangle is always} \\ \text{less than the third side.}
\displaystyle \textbf{For example:}
\displaystyle \text{In }\triangle ABC,\text{ given for Corollary 1, if }AB\text{ is the largest side and }AC\text{ is} \\ \text{the smallest side, then:}
\displaystyle \text{(i) }AB-AC<BC,
\displaystyle \text{(ii) }AB-BC<AC\text{ and}
\displaystyle \text{(iii) }BC-AC<AB.

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